
Copyright^ . 



COPYRIGHT DEPOSIT. 



DYNAMICS OF MACHINERY 






BY 



GAETANO LANZA, S.B., C. &M.E. 

Engineer; Professor Emeritus of Theoretical and Applied Mechanics 
Massachusetts Institute of Technology 



FIRST EDITION 

FIRST THOUSAND 



NEW YORK 
JOHN WILEY & SONS 

London: CHAPMAN & HALL, Limited 

1911 



A 



^ 



^ 



\r 



Copyright, 1911, 

BY 

GAETANO LANZA 



Stanhope JPress 

F. H. G1LSON COMPANY 
BOSTON. U.S.A. 

©CI.A297520 
o.V 



\\- 14/13 f 



PREFACE 



While Chapter I treats of the principal types of Dynamometers, 
the remainder of this book has for its chief object to bring together, 
in one volume, the methods of dealing with the inertia forces that 
arise in various kinds of machinery especially in cases where high 
speeds are employed. As examples, may be cited those of high- 
speed steam engines, including high-speed locomotives, and of gas 
engines. 

In these, careful consideration must be given to the action of 
the reciprocating parts, not only for the purpose of balancing, 
and hence avoiding undue strains in the machine itself, or in the 
foundations, and undue distortions in the rails, but also in order 
that the parts of the engine, including the crank shaft, etc., may 
be properly designed to resist the stresses to which they are sub- 
jected. 

Other examples in which the inertia forces must be given careful 
consideration are : the inertia governor, — inasmuch as these forces 
affect very considerably the regulation, — pulleys, flywheels, steam 
turbines, dynamo armatures, centrifugal machines, hydroextrac- 
tors, etc., which should be in running as well as in standing balance. 

Another set of examples includes those in which the gyro- 
scope is employed in engineering, as (a) the steering of torpedoes, 
(6) the steadying of vessels at sea, (c) the Brennan monorail car, 
(d) the gyroscopic compass, etc. 



in 



TABLE OF CONTENTS 

Chapter * Page 

I. Dynamometers 1 

II. Moments and Products of Inertia 20 

III. Action of Reciprocating Parts . 33 

IV. Governors 135 

V. Bodies having a High Rotative Spee ) 200 

Appendix to Chapter II (A) 223 

Appendix to Chapter IV (B) 232 

Appendix to Chapter V (C) 238 



DYNAMICS OF MACHINERY 



CHAPTER I. 

DYNAMOMETERS. 

Dynamometers, as their name implies, are instruments for 
measuring power. 

They may be divided into two main classes, viz., traction dyna- 
mometers and rotation dynamometers. The first are intended to 
measure the work done by a direct pull or thrust; as, for instance, 
the work done by a locomotive in drawing a train, or that re- 
quired to tow a boat. Rotation dynamometers, on the other hand, 
are intended to measure the power transmitted to or through a 
rotating shaft. 

Traction dynamometers are all practically some kind of a 
weighing device, the main part of which consists of a spring, or 
of a hydraulic cylinder and piston, by means of which the pull 
exerted is weighed, together with some device for measuring the 
speed of motion. 

When the pull and the speed are both constant, it is only neces- 
sary to multiply them together to obtain the work done per unit 
of time. On the other hand, when one or both vary, it becomes 
necessary to have recourse to some kind of a recording appa- 
ratus, and then to obtain the area of the resulting irregular 
figure by means of a planimeter or otherwise. 

Dynamometer Cars. 

Many of the large railroads make use of a dynamometer car, 
principally for the purpose of obtaining a tonnage rating, for the 
different parts of the service; determining the amount of energy, 
the draw-bar pull, and the power, and hence the kind of locomotive 
required to perform the service. 

In all these cases, the force with which the portion of the train 
behind the dynamometer car pulls upon the drawbar of the latter 
is weighed, and recorded upon a strip of paper, which is caused by 
suitable mechanism to travel at a speed proportional to that of 
the train. 

We thus obtain a diagram (see Fig. 1) in which the abscissae 
represent to scale distances travelled by the train along the road, 
while the ordinates represent to scale the draw-bar pull exerted at 

1 

Copyright, 191 1, by Gaetano Lanza. 



DYNAMICS OF MACHINERY 



the instant, and the area of the diagram included between any two 
given ordinates represents the work done during the interval in 
question. On the diagram are also recorded, by means of a 
chronograph, points usually five seconds of time apart. 




Fig. 1. 

By this means we are able to determine the average speed in 
any one five seconds, or in any longer interval. The longer the 
interval, however, the greater the accuracy. When five second 
intervals are used for computing the average speed, the probable 
error is very considerable. The method adopted for weighing the 
draw-bar pull is most commonly by means of a piston or plunger 
connected to the drawbar and moving in a closed cylinder of oil, 
the pressure of the oil in the cylinder being transmitted to a small 
indicator piston where the force is resisted by a calibrated spring, 
the motion of this small piston being multiplied by a suitable 
pencil mechanism by which the dynamometer diagram is drawn 
on the moving strip of paper. 

More or less other pencils are provided, as one to record the 
zero line of pull, one operated by hand to mark the mileposts, 
sometimes one to describe a speed curve, etc. 

One precaution taken in order to eliminate, as far as possible, 
errors due to the friction of the piston attached to the drawbar 
is to allow a certain amount of leakage around it. Also the 
constant jarring to which the piston is subjected by the motion 
of the train tends to minimize piston friction and hence to render 
this device more accurate than it would be if employed in a test- 
ing machine. Sometimes powerful springs are employed instead 
of a piston moving in a cylinder of oil. 

The train of mechanism that drives the strip of paper usually 
derives its motion from one of the truck axles of the dynamometer 
car. Sometimes, however, one of the trucks of the dynamometer 
car is provided with six wheels, of which the two middle ones are 
cylindrical instead of conical, and the paper-driving mechanism 
derives its motion from the axle of these cylindrical wheels. A 
diagram will be given showing the mode of applying the above in 
practical cases. 

In the case of the dynamometer car on the Pennsylvania Rail- 
road Mr. Emery (see Fig. 2) has constructed a more accurate 
device, which involves many refinements. 



DYNAMOMETERS 



s4jSi gm^/M\/n/i\/ft/flA\/-J/^ 




r^ 



P' 



tt 



Drawbar ' 





1 rf 



Fig. 2. 

Train Resistance. 

Another very important use of dynamometer diagrams, is for 
the determination of train resistance, i.e., its amount, and its 
variation with speed, number of cars, kind of cars, load carried, 
direction and force of the wind, curves, and other matters. This 
data would be especially useful in so designing locomotives that 
they may be able to perform the work required of them in the 
service for which they are intended. Thus far, however, but few 
attempts have been made to undertake a systematic set of tests 
which shall lead directly to a series of results of the character stated, 
and the conclusions reached in these regards have been arrived at 
from the results of a number of dynamometer tests made for 
other purposes, such as tonnage rating. Nevertheless there has 
been a very large amount of study of the subject of train resistance, 
arid various formulae have been deduced from such tests as were 
available, while a number of train resistance formulae have been 
proposed, and their relative merits discussed in articles that have 
appeared from time to time in the engineering periodicals. 



4 DYNAMICS OF MACHINERY 

Other Traction Dynamometers. 

Other uses for traction dynamometers are the determination of 
the energy and the pull required in towing boats, that of the energy 
and pull required in the case of road vehicles. Such a dynamom- 
eter should be calibrated before it is used, either by means of a 
testing machine or by applying weights directly. One form of 
traction dynamometer for use in towing boats will be shown in 
the following diagram. 




Dynamometers for Road Vehicles. 

While a number of dynamometers have been devised and 
used in the case of horse-drawn vehicles, there is but little call for 
them. In the case of electric cars and electric automobiles, the 
problem generally presents itself in a somewhat different form; 
viz.: It is necessary to ascertain the power that must be exerted 
by the motor to operate the car at the speed, and under the con- 
ditions existing in service. In these cases suitable electric meas- 
uring instruments are employed. In the case of gasoline or of 
steam-driven automobiles, it is usually impossible to take indi- 
cator cards, and it becomes necessary to measure the energy de- 
livered to the driving shaft by means of some form of rotation 
dynamometer. The input can then be determined by means of 
a series of laboratory tests of the motor, and thus the amount of 
gasolene, or of oil, or the B.t.u.'s can be ascertained. 



Transmission Dynamometers. 

Transmission dynamometers include those that are designed to 
measure, but not to consume. 

(a) The power transmitted from one revolving shaft to another 
with which it is connected by suitable mechanism, as 
belt gearing, toothed gearing, etc. 

(6) The power transmitted through a revolving shaft. 

By way of illustration, suppose we wish to determine the power 
required to drive a machine, as a spinning frame or a lathe; when 
performing their ordinary work we may interpose a transmission 
dynamometer of kind (a) between the machine and the counter- 
shaft from which it is driven, and instead of driving the machine 



DYNAMOMETERS 5 

from the shaft we drive the dynamometer from the shaft and the 
machine from the dynamometer, thus transmitting the power 
through the dynamometer, and weighing it on its passage. On the 
other hand, suppose we wish to ascertain the power transmitted 
through the propeller shaft of a steamer, we attach to the shaft a 
dynamometer of kind (b), and thus weigh the power transmitted 
through the shaft to the propeller. 

Transmission dynamometers are generally constructed on one 
of the two following principles, viz. 

(1) The driving moment (which is the product of the force 

exerted to drive, and its lever arm) is weighed. The 
number of revolutions per minute or per second being 
also observed, the power transmitted is readily com- 
puted as follows : 

Let M = driving moment in foot-pounds. 

N = number of revolutions of shaft per minute. 
H.P. = horse power transmitted. 

Then HP = 3P00- 

(2) The angle of twist of the shaft is measured, and from this 

is determined the driving moment. The number of 
revolutions per minute or per second being also ob- 
served, the power transmitted is readily computed. 

Of those built on the first principle we may distinguish two 
classes, viz.: 

(a) Those in which the force exerted is weighed by levers and 
weights. 

(6) Those in which the force exerted is weighed by a spring. 
They will be illustrated by several examples, as fol- 
lows: 

Differential Dynamometers. 

In these dynamometers levers and weights are used to deter- 
mine the power transmitted, but they also contain an epicyclic 
train of four bevel gears. While this type was originally devised 
by Samuel Batchelder, minor modifications for convenience in 
use have been made by others. The following cut shows in out- 
line, the form adopted by Webber and by Silver and Gay. 

The dynamometer receives its motion from a belt connecting 
the pulley on the countershaft, or other source of power, with the 
pulley A, and transmits it by means of a belt connecting the 
pulley B with that on the machine whose power is to be deter- 
mined. Pulley A and spur gear D are fast on shaft C, while spur 
gear E, and bevel gear F are fast on sleeve G which is free to 
revolve around shaft H, this latter revolving in the bearings KK. 
Moreover bevel gear L and pulley B are fast on shaft H. The 



DYNAMICS OF MACHINERY 

motion received by pulley A is transmitted through shaft C, spur 
gear D, spur gear E, and sleeve G to bevel gear F. Between bevel 
gears F and L are inserted the pair of bevel gears M and N which 
are loose upon the cross shaft 0, through which shaft H passes 
freely. An extension P of this cross shaft forward forms a gradu- 
ated scalebeam, at the end of which is a knife-edge from which 




Fig. 4. 
< 
hangs a scalepan. A short extension Q of the cross shaft back- 
wards furnishes the means for attaching an adjustable counter- 
weight and a dashpot. If the cross shaft were not held in place 
by some outside force the bevel gear L and pulley B would not 
turn, but the cross shaft would revolve at one half the speed 
of bevel gear F. But if the cross shaft is held in place by some 
outside force or if it is locked in place, then the motion of pulley A 
and hence of bevel gear F will cause bevel gear L and hence pulley 
B to revolve at the same speed as pulley A. 



Mode of Graduation. 

Just as with any ordinary scale there are two means of weigh- 
ing provided, viz.: one by putting weights in the scalepan, and 
the other by means of a sliding poise on the scalebeam. More- 
over we are at liberty to assume arbitrarily, (a) the position of the 
knife-edge from which hangs the scalepan, (b) the zero position 
for the sliding poise, i.e., that corresponding to zero load, and 
(c) the position corresponding to any given load as 1000 foot- 
pounds per 100 revolutions, when the scalepan is empty; provided 



DYNAMOMETERS 7 

(a) we determine the weight of the sliding poise to correspond to 
the values chosen, (|3) we counterweight correctly the dynamometer 
at rest, with the sliding poise at zero, and the scalepan empty; 
and (7) we determine the weights in the scalepan which will 
correspond to a given load as to 1000 foot-pounds per 100 revolu- 
tions. 

As to the weights in the scalepan, Webber so locates the knife- 
edge from which hangs the scalepan, that this knife-edge would 
if left free to turn describe a circle having a circumference of 
ten feet and since the lever would in that case make one half the 
number of revolutions that the shaft makes, it follows that two 
pounds in the scalepan would correspond to 1 X 10 = 10 foot- 
pounds per revolution of the shaft, or to 1000 foot-pounds per 100 
revolutions. Hence the 2-pound weights would be marked 1000. 

Next as to the sliding poise. Assume 

Fig. 5 that a is the center of the system of y ° & c 

bevel gears, that c is the position of the [J 

knife-edge from which hangs the scale- 
pan, that o is the arbitrarily chosen zero 
point for the counterpoise, and b the point jrj g 5 # 

where it is to register 10 foot-pounds per 
revolution 

Let ob = d in feet. 

Let x = weight of sliding poise in pounds. 

Observe that the effect of placing the sliding poise at b instead 
of o is the same as the effect of leaving the sliding poise at o, 
and placing a 2-pound weight in the scalepan. 

Then take moments about (a) , and we have 

x (ab) = x (ao) + 2 (ac), .'. x (ab — ao) = 2 (ac). 
But ab — ao = ab = d, and in the Webber dynamometer 

10 , , 10 10 

ac = — leet. .*. xa = — , .*. x = — 5 

Ztt 7T TTCi 

In the Webber dynamometer d = — feet. 

12 
.'. x = — = 3.818 pounds, .*. x = 3 pounds 13 ounces 1.4 dr. 

The divisions on the scalebeam are to be made of equal lengths. 

Emerson Power Scale. 

Another transmission dynamometer of the first kind is the 
Emerson Power Scale, which is shown in outline in Figs. 6 and 
7. To explain its construction and action, suppose we wish to 
ascertain the power required to drive a spinning frame, or a ma- 
chine tool, we may proceed as follows, viz. : 



8 



DYNAMICS OF MACHINERY 



Remove the loose pulley from the frame, or machine tool ; 
loosen the set screws of the tight pulley, and file off the burr made 

by the set screw so that the 
tight pulley is made loose 
upon the shaft of the frame, 
and is in the same place 
where it was before. 

Now fix the dynamometer 
fast on the shaft in such a 
position that the prongs A A 
project between the arms of 
the pulley. See to it that 
both prongs bear equally 
against an arm. This may 
be accomplished either by 
means of shims, or else by 
filing one of the arms at the 
proper place. Then when 
the belt is put on, the pulley 
is driven by the belt and the 
pulley drives the outer ring 




SECTION XY 

Fig. 6. 



B of the dynamometer. This outer ring transmits motion to the 
spider of the dynamometer 



bQdd 



C, thence to the shaft D, 
and hence to the frame 
through the levers EFG, 
which are attached by pins 
to the ring B at E, and to 
the spider by pins at F. 
Assume the rotation to be 
right-handed. Then the 
rods GH push outwards 
upon the ends H of the 
short arms KH of the 
angle levers HKI, whose 
fulcrums are at K. This 
causes the ends I of the 
long arms of the angle 
levers to move to the right 
in Fig. 7. 

From the ends of the 
long arms of the angle 
levers project two rods IL 
which are fastened at L 
to a collar N, which ro- 
tates with the instrument, 
and is free to slide on the 
hub of the dynamometer. 
Set in a groove in the collar N is another collar jlf, which 




Fig. 7. 



DYNAMOMETERS 9 

does not rotate, but which is moved longitudinally along the 
shaft by the collar N. This collar M is attached to one (forked) 
end of angle lever R. This carries at its other end a scalepan T, 
and also moves a pointer P over a graduated scale W. While the 
weight in the scalepan required to indicate a certain power can be 
calculated roughly from the leverages, it is necessary in order 
to obtain satisfactory results to calibrate the instrument by com- 
parison with a brake. 

Of course the tare must be ascertained in each experiment, 
and subtracted from the reading* 

The shaft under the pulley should be made smooth, and should 
be oiled, or better, the pulley should be provided with a ball bear- 
ing to minimize its friction. 

Van Winkle Dynamometer. 

The Van Winkle dynamometer operates in a way similar to the 
Emerson Power Scale. Its construction differs mainly in the fact 
that instead of a ring connected with the spider by levers, whose 
motion is ultimately resisted by placing weights in a scalepan, 
hung from a knife-edge at the end of the system, two parallel disks 
A and B are employed, connected by two coil springs, and the 
amount of stretching of the springs, and hence the corresponding 
angle of turning of one disk by the other, is caused by suitable 
mechanism to move a sliding collar F along the shaft, and the 
amount of this sliding is indicated on a graduated dial H placed 
at the end of the system. 

The following cut shows the dynamometer in outline. 



^^^"^^^^^ 



^^^ 




Fig. 8. 



The Belt Dynamometers. 

The main objection to all belt dynamometers, is the error in- 
troduced by the slip of the belt. The same objection holds in a 
lesser degree in the case of all dynamometers where a belt is used 



10 



DYNAMICS OF MACHINERY 




Fig. 9. 



to connect them with either the source of power, or with the 
machine whose power is to be measured. Only one belt dyna- 
mometer will be described here, viz., the Tatham. 

Tatham Dynamometer. 
The Tatham dynamometer may be called a belt dynamometer. 
The general action is shown in Fig. 9. 

The power applied to the shaft on which the driving pulley D 
is fixed is transmitted to the pulley B, the shaft of pulley B 

being coupled to the machine (the power 
to drive which is to be ascertained) by 
an endless belt which passes over D under 
the stretching pulley S, over the weighing 
pulley W, under B, over the second weigh- 
ing pulley W, under S, back to the place 
of starting. Each of the weighing pulleys 
is supported in a cradle, the outer end of 
which is pivoted on the knife-edge F y 
while the inner edge is supported by the 
link LC. The upper ends of the two links 
are fastened to the scalebeam FiP at 
equal distances from and on either side 
of the fulcrum F\. 
To calculate the power applied to the pulley B it is necessary 
to know three things: the difference of tensions of the belt on the 
two sides of B, its effective diameter, and its number of revolutions. 
The scalebeam is acted on through the links LC fastened to the 
cradles of the weighing pulleys W. The tensions of the belt on 
the outer faces of these pulleys have no effect on the beam, since 
the line of effort of the belt passes through the knife-edges F. 
The only forces that act on the beam are, therefore, the two ten- 
sions of the belt on the inner faces of the pulleys W, and these 
are the tensions on the two sides of B; and the links, being at equal 
distances on either side of F, the difference of the tensions is 
recorded on the beam. A counter records the number of revolu- 
tions, and we thus have the means of determining the power 
transmitted. 

Cradle Dynamometer. 

Another form of dynamometer 
which has been used to some 
extent to measure the power of 
small dynamos and other small 
machines is the Cradle dyna- 
mometer. 

It consists of a cradle hung on 
two knife-edges at H and G, 
Fig. 10, C in the side elevation. 



g- 




Fig. 10. 



Hence it is free to rock about the line HG (C in the side elevation). 



DYNAMOMETERS 11 

The movable weights D enable us to bring the center of gravity 
of the dynamometer, with the machine on it, to C. 

There is also a scalebeam AA\ attached rigidly to one side of 
the cradle, preferably near C. 

The manner of using it, is as follows : 

Place the machine whose power is to be determined in the 
cradle and so adjust it that the axis of the driving pulley shall be 
in line with the knife-edges. Adjust the movable weights D so 
that the center of gravity of dynamometer and machine combined 
may be in the line HG. Place weights in the scalepan of sufficient 
amount to bring the platform BE horizontal or the rod CD verti- 
cal. Now belt onto the driving pulley from outside. This belt 
should preferably be vertical, and the driving should be from 
below, as otherwise the errors are large. Moreover, adjust the 
direction of motion of the machine driven so as to tend when 
run to tip the dynamometer in such a direction as to raise the 
scalebeam AA\. Then when the machine is running bring the 
scalebeam back to a horizontal position by means of weights in 
the scalepan. Then the moment of the weight on AA\ about C is 
just equal to the product of the difference of tensions in the driv- 
ing belt by the radius of the pulley, and this multiplied by 2 tt 
times the number of turns per minute gives the work done per 
minute. 

Torsion Meters. 

If power is transmitted through a shaft whose shearing modulus 
of elasticity is known, we can determine the power transmitted 
by measuring the angle of twist of a certain gauged length, and the 
number of revolutions per minute, as follows: 

Let r — outside radius of shaft in inches. 

ri = inside radius of shaft in inches. For a solid shaft r\ = 0. 

G = shearing modulus of elasticity in pounds per square inch. 

L — gauged length in inches. 

a = angle of twist in gauged length L in radians. 

N = number of revolutions per minute. 

7r (r 4 — ri 4 ) 
I = — - — = polar moment of inertia of section in 

(inches) 4 . 
H.P. = horse power transmitted. 

,, (33,000 H.P.) 12 . ,. , . , , 

M = _ %T = twisting moment in inch-pounds. 

ZttJS 

We then have 

ML ,, aGI , . TTT . 2wNM 



a 



M = -^r— and since H.P. = 



GI " L * (12) (33,000) 

we have 

HP = 2 ^ x g «L 
12 X 33,000 L 



12 



DYNAMICS OF MACHINERY 



Example. — Given a solid shaft 12 inches diameter, for which 
G = 12,000,000, and N = 80. Suppose that by means of the 
torsion meter we find a = 0.06 radians for a gauged length of 
1000 inches. Find the horse power transmitted. 

Solution. — In this case we have 

r = 6, n = 0, G = 12,000,000, L = 1000, a = 0.06, N = 80, I = ^^ • 



Hence 
H.P.= 



2tt(80) 



(0.06) 



12,000,000 ttX 1296 



12 X 33,000 

7T 2 (80) (0.06) X 1296 
33 



1000 X 2 

(9.87) (80) (0.06) (1296) 
33 



= 1860. 



Dynamometers which determine the power transmitted by meas- 
uring the angle of torsion of a shaft in a certain gauged length are 
called Torsion Meters. 

The first dynamometer of this kind was the Pandynamometer of 
Hirn, a rather crude instrument. 

Later quite a number of torsion meters have been devised, and 
their principal application has been and is the determination of 
the power transmitted through the propeller shaft of a boat. 
One of the first of these was devised by Denny and Johnson, and 
in it electrical means were employed to determine the angle of 
twist. Moreover, it is not a recording dynamometer. Later we 
have those of Hopkinson, of Frahm, and of Fottinger. 

A brief description of the last will be given. It employs mechan- 
ical means for finding the angle of twist. It is also made record- 
ing, and by its use we can determine , not only the mean angle of 
twist of the shaft, but also its vibrations. 



Recording Cylinder 



77777777. 




Fig. 11. 

A sleeve (a) is keyed, or otherwise fastened to the shaft at (6) ; 
the other end (c), and the disk (d) which is fastened on the sleeve, 
are free, but all are so adjusted as to be central on the shaft. At 
(e) is another disk fastened rigidly to the shaft. Hence the 
angular displacement observed in the two disks (d) and (e), when 
power is transmitted, determines the angle of twist in the test 
length as shown in Fig. 11. 



DYNAMOMETERS 13 

The sleeve, the two disks, and lever systems, revolving with the 
shaft, are found in most of the Fbttinger meters ; on the other hand, 
there is more or less variety in the different devices adopted to 
obtain scale readings, or self-recorded diagrams, according to 
the exigencies of the case. The lever system will be evident from 
the figure. In the case of the one shown, a recording drum is 
added, which does not rotate with the shaft, and the stylus 
operated by the system traces a curve, whose ordinates indicate 
the angle of twist at any instant. By using a suitable scale they 
can, however, be made to indicate the tangential effective force at 
any instant, which when multiplied by the radius of the shaft 
gives us the twisting moment; from which, when the number of 
revolutions per minute is known, the horse power can be readily 
determined. 

Moreover, the drum can be pushed to a new position along the 
shaft, in order to obtain a record for a different revolution, or, it 
can be pushed so far along the shaft as to be free of the stylus, 
when a new paper can be mounted upon it. When the shaft diam- 
eter is large, then record sheets become unhandy, and, instead of 
the drum, a spur gear is mounted on the shaft, which by engaging 
with another spur gear drives a drum riding on the shaft. 

One important matter is the determination of the shearing 
modulus of elasticity of the shaft. This can be done by using the 
apparatus to determine the angle of twist, when the shaft is at 
rest, by applying known twisting moments through weights, or 
otherwise. 

Absorption Dynamometers or Brakes. 

The object of an absorption dynamometer, otherwise called a 
brake, is to absorb the power transmitted by a shaft, and, at the 
same time, to measure it. 

Brief descriptions will be given of some of the main types, and 
the principles underlying their mode of action will be explained. 

Prony Brakes. 
The most essential parts of a Prony brake are: 

1° A band embracing the pulley, which can be tightened so 
as to develop the amount of friction required. 

2° A stream of cold water, circulating either through the 
band, or through the rim of the pulley, to absorb and 
carry away the heat developed by the friction. 

3° A lever arm, with a knife-edge at its outer end. 

4° Means for applying a force at this knife-edge, in the proper 
direction to weigh the power. This may consist of a 
scale pan in which weights are placed, or it may con- 
sist of a platform scale or of a spring. 

5° A revolution counter, to determine the number of revolu- 
tions per unit of time. ' 



14 



DYNAMICS OF MACHINERY 



Thus, the brake shown in Fig. 12 has a band which can be 
tightened by a screw B. In this case the water circulates through 
a pipe in the rim of the pulley, the rotation of the latter being in 
the direction shown by the arrow. The tendency of the pulley 
is to rotate the entire brake band and lever in the same direction, 
but this is prevented, and the pulley alone is left to rotate, in 
^consequence of the weights in the scalepan E. In such a brake 
the brake band, lever, and scalepan should be independently 
counterbalanced by hanging them, at their center of gravity, to a 
counterweight ing lever, as shown in the cut, as otherwise there will 
be a tare reading to be taken into account. 




Fig. 12. 

The work done when the brake is counterweighted as described 
above is found by multiplying the weight in the scale pan by the 
circumference of a circle whose radius is OA, and by the number 
of revolutions of the pulley. Thus, suppose OA = 63.02 inches. 
The circumference of the circle whose radius is OA is, in feet 



2 7T (63.02) 
12 



= 33 feet. 



and, if the load in the scale pan be 5 pounds, and the number of 
revolutions per minute be 200, then the work done per minute is 

5 X 33 X 200 = 33,000 foot-pounds, 

i.e., 1 horse power. If the brake is not counterweighted, there 
should be added to this the product of the combined weights of 
lever and scale pan by the circumference whose radius is the dis- 
tance from to their center of gravity multiplied by the number 
of revolutions per minute. 

There are various other devices for a brake band, and for the 
means of tightening it, as well as for the weighing of the load. 
Some of these are shown in the following cuts, which explain them- 
selves. 



DYNAMOMETERS 



15 



B B 

d°i a. 




fo 




w 



Fig. 13. 



t°t 



6 w 



Fig. 14. 




w$///Ms////M/bti&& 



Fig. 15. 




Fig. 16. 



16 



DYNAMICS OF MACHINERY 



On the other hand, it is to be observed that, whenever we use 
only one lever, any weight placed in the scalepan will increase 
the load on the journal, and hence cause the friction in the boxes 
to vary. 

This source of error can be avoided by using two lever arms, 
one at each end as shown in Fig. 17, and loading one upwards 




Fig. 17. 

while the other is loaded downwards, these two loads forming a 
statical couple; thus the load on the boxes and the friction in the 
boxes are constant. The upward weight can, if desired, be applied 
by means of a spring. Fig. 18 shows a brake in which a rope is 

attached to a wooden horse CB 
resting on a platform scale, is 
wound around the pulley 0, and 
fastened to a hook supported by 
B, vertically over one in C, the 
pulley being so located that the 
directions in which the rope 
leaves the pulley, whether up- 
ward or downward, are vertical. 
The rope is tightened by a hand 
wheel and screw at H, by means 
of which the hook can be raised 
g- 8 - or lowered. 

When using a dynamometer similar to Fig. 15 the tare reading 
may be found as follows : Support the brake by a cord $ directly 
over the center of the shaft; after loosening the brake band by 
the screw at B, then note the scale reading. This reading should 
be subtracted from all scale readings when calculating the power. 




DYNAMOMETERS 



17 



Automatic Devices. 

There are various automatic 
devices for keeping the load con- 
stant. One of them is shown in 
Fig. 19. 

If for any reason the friction 
increases, its very increase lifts 
the lever with the small weight 
w, and causes the friction to de- 
crease, and vice versa. 

Alden Brake. 




Fig. 19. 



Water Outlet 



The Alden brake differs from a Prony brake in the facts that no 
band is employed, but that the friction is developed between a 

series of copper plates and a set 
of cast-iron disks, pressed to- 
gether by water pressure on the 
outside of the copper plates, 
while the rubbing surfaces are 
thoroughly lubricated. More- 
over, the area of contact being 
large, the water pressure required 
is moderate. 

This brake consists of 

(a) A series of revolving cast- 
iron disks EE keyed to 
the shaft which trans- 
mits the power. 

(6) A non-revolving casing B 
having its bearings on 
the hubs A of the re- 
volving cast-iron disks. 

(c) A pair of copper plates C in 
contact with each cast- 
iron disk, these plates 
being fastened to the 
casing. Moreover, each 
cast-iron disk with its 
two copper plates forms 
one unit. 

brake is 




Water Outlet 



Fig. 20. 



When the oraKe is m use, 
water under pressure fills the 
spaces between the units and 
between the end units and the casing, and its pressure tends to 
force the plates against the disks. The greater this pressure, 
the greater the friction between plates and disks. 



18 



DYNAMICS OF MACHINERY 



The casing and plates are prevented from being made to revolve, 
with the disk, by means of the weighing apparatus shown in Fig. 21 ; 
i.e., the weights counteract the tendency of the casing to revolve 
with the disks. 




VMS5-J~?<S,-<\-s^^^^^/A.^^\Wt,W^ 



Fig. 21 



Another system of piping circulates oil for lubricating the sur- 
faces of contact between disks and plates. 



Air-resistance Brakes. 

In these brakes, the power is absorbed by the resistance of the 
air to the motion of a set of vanes on a revolving shaft. 

This class of dynamometer will be illustrated by the following 
example. 

Renard Dynamometric Fan. 

This apparatus was designed, and has been employed instead 
of a brake, for the measurement of the power of high-speed small 
motors. 

To the end and at. right angles to the shaft of the motor is firmly 
attached a bar of wood or metal, which carries two vanes placed 
at equal distances from the center of the shaft. The power 
developed by the motor is absorbed by this fan, and, after the 
latter has been calibrated experimentally, so that the power re- 
quired to drive the fan at different speeds has been ascertained, 
and the results have been plotted, or tabulated, it is only neces- 
sary to observe the speed of the shaft, and then to determine from 
the plot or the table the power transmitted. The calibration is 
affected by means of a device called a dynamometric balance, 
which is similar in principle to a cradle dynamometer, and which 
may be described as follows: 

An equal armed scale beam, provided at its ends with scale 
pans SS, is mounted at its center on a knife edge. To and above 



DYNAMOMETERS 



19 



the scale beam is firmly attached a frame which carries the motor 
M and its fan VV. Hanging from the center of the frame is a 
rod, which carries an adjustable weight C for the purpose of rais- 
ing or lowering the center of gravity of the entire apparatus. 
Also a needle and a graduated arc 
are provided to show when the 
scale beam is horizontal. 

Before motion is imparted to the 
fan, the weights in the scale pan 
are so adjusted as to bring the 
pointer to zero, thus balancing the 
apparatus. The current is then 
turned on, and when the desired 
speed is reached suitable weights 
are added in one of the scale pans, 
until the pointer is again brought 
to zero. 

The product of this added weight 
by its leverage about the knife 
edge is the turning moment, and 
the product of this by 2ir times 
the number of turns per minute 
is the work done per minute. 

The horse power is obtained by 
dividing this last product, if it is in foot-pounds, by 33,000, or 
if in inch-pounds, by 33,000 X 12. 




Fig. 22. 



Electric Dynamometers. 

The entire subject of the electric measurement of power is 
foreign to the purposes for which this book is written. 

Sometimes the power is measured without any apparatus that 
could properly be called a dynamometer. 

On the other hand, when the power is to be absorbed electrically 
as well as measured, it may be done by means of a water rheostat 
provided with suitable electrical measuring instruments. 

Again, we sometimes have apparatus that may properly be 
called an electric dynamometer; as, for instance, when we con- 
struct an apparatus similar to a Prony brake, except that instead 
of a band for the brake wheel we use the armature of a dynamo. 

In still another kind, we may have a mechanical dynamometer 
with some electrical attachment. 



CHAPTER II. 

MOMENTS AND PRODUCTS OF INERTIA. 

If we refer a body to three rectangular axes OX, OY, and OZ 
(Fig. 23), then, by definition, we have for the moments of inertia 
about these axes, 

A = limit of 2w (y 2 + z 2 ), about OX; 
B = limit of 2w (x 2 + z 2 ), about OY; 
C = limit of 2w (x 2 + y 2 ), about OZ. 




Fig. 23. 

For the moments of inertia with reference to the coordinate 
planes, we have 

A' = limit of liWx 2 , with respect to the X plane, i.e., the plane YOZ; 
B' = limit of Hwy 2 , with respect to the Y plane, i.e., the plane XOZ; 
C = limit of 2it>z 2 , with respect to the Z plane, i.e., the plane XOY. 

For the moment of inertia with respect to the origin 

H= limit of 2w (x 2 + y 2 + z 2 ). 

For the products of inertia 

D = limit of Xwyz; E = limit of Xwxz; F= limit of Xwxy. 

Moment of Inertia about any Axis. 

For the moment of inertia about an axis OY through the origin, 
which makes with the axes OX, OY, and OZ respectively, angles 

XOV=a, YOV=p, and ZOV=y, 
20 



MOMENTS AND PRODUCTS OF INERTIA 21 

we shall have 

I = A cos 2 a -f- B cos 2 fi + C cos 2 7 — 2 D cos /3 cost— 2 E cos a cosy 

— 2F cos a cos 13 (1) 

Proof. — From any point P of the body whose coordinates are 
x, y, and z, draw a perpendicular PQ to the axis OV. 
Then 

0P2 = X 2 + ^2 _|_ ^ 

OQ = projection of OP on OF 

= projection of broken line OABP on OF 
= projection of OA on OV + projection of A B on OV 

+ projection of BP on OF. 
.*. OQ — x cos a + 2/ cos ]8 + 2 cost; 

.*. QP 2 = OP 2 — OQ 2 = x 2 -\- y 2 -\- z 2 — (x cos a + y cos /3 + 2 cos t) 2 ; 
/. QP 2 = x 2 (1 - cos 2 a) + ?/ 2 (1 - cos 2 /3) + z 2 (1 - cos 2 t) 

— 2yz cos |S cosy — 2 a:^ cos a cosy 

— 2 xy cos a cos 0. 



But since 
we have 



COS 2 a + COS 2 |8 + COS 2 Y = 1, 



1 — cos 2 a = cos 2 /3 + cos 2 y, 1 — cos 2 /3 = cos 2 a + cos 2 y, 
1 — cos 2 y = cos 2 a + cos 2 /3. 
.*. QP 2 = (^/ 2 + z 2 ) cos 2 a + (x 2 + 2 2 ) COS 2 |8 + (x 2 + t/ 2 ) COS 2 Y, 

— 2yz cos |8 cos y — 2 ^ cos a cos 7 — 2 £?/ cos o: cos /3 ; 
and since 

7 = limit of Xw'QP 2 
we have 

I = A cos 2 a + P cos 2 18 + C cos 2 y — 2D cos /3 cos 7 

— 2 E cos a cos 7 — 2 P cos a cos/3. Q. E. D. 

Moments of Inertia about Parallel Axes. 

The moment of inertia of a body about an axis not passing 
through its center of gravity is equal to its moment of inertia 
about a parallel axis passing through its center of gravity increased 
by the product of its entire weight by the square of the distance 
between the two axes. 

Proof. — Refer the body to a system of three rectangular axes 
OX, OY, and OZ, and let OZ be the one about which the moment 
of inertia is desired. 

Let the coordinates of the center of gravity 0\, of the body with 
reference to OX, OY, and OZ, be x , yo, z . With 0\ as an origin 
of coordinates construct a new system of axes 0\X\, 0\Y\, and 
0\Z\ respectively parallel to OX, OY, and OZ. Let the coordi- 
nates of the point P with reference to these new coordinate axes 
be X\, y\, Z\, then we have 

X = Xi + x Q , y = y l + y , and z = z\ + z . 



22 DYNAMICS OF MACHINERY 

Let 

W = limit of 2w = entire weight of the body. 
Let 

Co = limit of 2w (xi 2 + y^) = moment of inertia about OiZ h 
and 

C = limit of Xw (x 2 + y 2 ) = moment of inertia about OZ. 

Let 

r = Vxo 2 + 2/0 2 = perpendicular distance between OZ and 0\Z\. 
Then we have 

C = limit of 2w (z 2 + y 2 ) = limit of 2mr 2 + limit of Xwy 2 . 
Hence 

C = limit of 2w (xi + x ) 2 + limit of 2w {y x + 2/o) 2 , 
= limit of Dwzi 2 + limit of 2wyi 2 + x 2 limit of 2w, 
+ i/o 2 limit of 2w + 2 x limit of 2w;xi + 2 1/ limit of 2wyi. 

But since Xi, 2/1, zi, are the coordinates of any point with reference 
to OiXi, OiYi, and OiZ h and since 0\ is the center of gravity, we 
have 

limit of 2wxi = 0, limit of Styj/i = 0, 
/. C = limit of 2w (^! 2 + y x 2 ) + TF(x 2 + y 2 ) 
... c = C + TF Oo 2 + 2/o 2 ) = C + TFr 2 . Q. E. D. 

Principal Axes of Inertia. 

It will be shown in the Appendix that, for any given point taken 
as origin, there exists at least one system of three rectangular 
axes, for which the products of inertia of the body are zero, and 
that the moments of inertia of the body about these axes fulfill the 
conditions required to render them either maximum or minimum; 
one of the three being a maximum, and one a minimum, while 
the third, which is numerically intermediate between the other 
two, is a maximum when compared with other axes lying in certain 
planes containing this axis, and a minimum when compared with 
other axes which lie in other planes containing this axis. 

Principal axes of inertia are most commonly defined as those 
for which the products of inertia are zero. Thus, were OX, OY, 
OZ, the principal axes, and hence A, B, and C, the principal mo- 
ments of inertia then would D = E = F = 0, and equation (1), i.e., 

I = A cos 2 a + B cos 2 /3 + C cos 2 y — 2D cos |8 cos y 
— 2 E cos a cosy — 2 F cos a cos /3, 

would become 

I = A cos 2 a + B cos 2 /3 + C cos 2 t (2) 

Principal axes of inertia are also defined as those about which 
the moments of inertia fulfill the conditions required in order that 
they may be either maximum or minimum. 



MOMENTS AND PRODUCTS OF INERTIA 



23 



These two definitions are equivalent, as will be shown in the 
Appendix, when methods will be developed for determining the 
positions of the principal axes, and the values of the principal 
moments of inertia. 

The work, however, involved in applying the results, in their 
general form, to cases that arise in engineering is, in most instances, 
unduly long, and an easier method is to employ the results of the 
solution of the following problem. 

Problem. — Given three rectangular axes, OX, OY, and OZ, 
Fig. 24, together with the moments and products of inertia about 
these axes, to find a point M on OZ if possible, which, being taken 
as a new origin, shall cause OZ to be a principal axis, and to find 
the other two principal axes MXi and MYi, which are, of course, 
perpendicular to M Z, 




Fig. 24. 

Solution. — Draw MX' and MY' parallel to OX and OY re- 
spectively. LetOM = d, and let angle X'MX X = angle Y'MY X = 
3 . The problem then reduces to finding the values of d and 3 . 
Let P be any point of the body, whose coordinates with reference 
to OX, OY, and OZ are x, y, and z. Let the coordinates of P with 
reference to MX h MY h and MZ be x h y\, and z\ .'. Ma' = x h 
Mb' = yi, P\P = z\ = z — d. We shall then have from the 
figure, 

xi = Ma' = projection of MP X on MX X 

= projection of broken line MaPi on MXi = x cos 3 + y sin 3 , 
y x = Mb' = projection of MP X on MY X 

= projection of broken line MbP\ on MYi = y cos 6 3 — x sin 3 . 
.*. Xi = x cos 3 + y sin 3 , yi = y cos 3 — x sin 3 , z\ = z — d. 

From these we obtain 

2/iZi = yz cos 3 — xz sin 3 — d (y cos 3 — x sin 3 ), 
XiZi = yz sin 3 + xz cos 3 — d (y sin 3 + x cos 3 ), 
£i2/i = (y 2 ~ x 2 ) cos 3 sin 3 + xy (cos 2 3 — sin 2 3 ). 



24 DYNAMICS OF MACHINERY 

Hence if we let 

Di = limit of ZwyiZi, 

Ei = limit of XwxiZi, 
and 

Fi = limit of Xwxit/i 
and observe that 

D = limit of Hwyz, 

E — limit of Xwxz, 
and 

F — limit of Xwxy, 

W = limit of 2w , 
x W = limit of Xwx, 
and 

y W = limit of 2wy, 
we obtain 

Di = D cos 03 — E sin 3 — dy W cos 3 + rfx TF sin 3 , 
Ei = D sin dz-\- E cos 3 — cfa/oT^ sin 3 — ^ TT cos 3 , 
Fi = (A - B) sin 3 cos 3 + F (cos 2 3 - sin 2 3 ) 

A — B 
= — ^ — sin 2 3 + F cos 2 3 . 

Since, however, ilOTi, MFi, and ilfZ are principal axes, we must 
have Di = Ei = Fi = 0. Hence we have the three equations: 

D cos 3 — E sin 3 — dy W cos 3 + dx W sin 3 = 0, 
D sin 3 + E cos 3 — dy W sin 3 — dx W cos 3 = 0, 
(A -B)sin2 03 + 2Fcos2 3 = 0. 

Solving the last equation for 3 , we have 

tan 2 3 = -j^zrj ( 3 ) 

Multiplying the first by cos 3 and the second by sin 3 , and adding 
D-dy W=0. .: d = ^ (4) 

Multiplying the first by sin 3 , and the second by cos 03, and sub- 

E 
tracting E — dx W = 0. .*. d = —^z (5) 

Equating these values of D, we have Dx = Ey , and this is the 
condition that it may be possible to find a point M on OZ which, 
being taken as a new origin, shall cause OZ to be a principal axis 
atikf. 

As a special case, observe that when D = E = 0, equations (4) 
and (5) are both satisfied by d = 0. 

Hence when D = E = 0, OZ is a principal axis at 0; 

similarly when D = F = 0, OF is a principal axis at 0, 
and when E = F = 0, OX is a principal axis at 0. 



MOMENTS AND PKODUCTS OF INERTIA 25 

If, moreover, for a given point M, on OZ, it is known that OZ is 
a principal axis at M, then the positions of MX\ and MY\ can be 
found from the equation 

tan2 3 = ^Z~J ( 6 > 

If, for a given point Mi on OX, it is known that OX is a principal 
axis at M h then the positions of ikfiFi and M\Z\ can be found 
from the equation 

tan 2 0! = ^g, (7) 

where 0i = angle made by M\Yi with a line through M\ parallel 
to OY. If, for a given point M 2 on OF, it is known that OF is a 
principal axis at M 2 , then the positions of M 2 X 2 and M 2 Z 2 can be 
found from the equation 

2 W 
tan 2 6> 2 = , _ ^ , (8) 

where 2 = angle made by M 2 X± with a line through M 2 parallel 
to OX. 

A further discussion of the Principal Axes and Principal Moments 
of Inertia will be found in the Appendix. 

Moments and Products of Inertia about Various Axes. 

In order to facilitate the solution of examples, a number of mo- 
ments and products of inertia that may be found convenient for 
use in certain applications to engineering purposes, will be given. 
In all these cases the following notation will be used: 

p = weight per unit of volume; 
V = total volume of the body; 
W = total weight of the body = pV. 

I. Sphere. 
Diameter = d. Axes of coordinates passing through the center. 
A= B = C = ^ Wd\ 



II. Homogeneous Circular Cylinder. (Fig. 25.) 
Outside diameter = d\, inside diameter = d 2 ; length = I. 

ird 2 

1° Solid Cylinder. — In this case d 2 = 0, V = —r— I. 
(a) Axis AB as axis of z, origin at 0. 

A = B = w sdi+ A1 . c = wf. 

48 8 



26 



DYNAMICS OF MACHINERY 



(b) Axis AB as axis of z, origin at C, where 
OC = a. 

A = B==W \ 6 ^ + 4 * +a2 . c = H 



48 



8 



(c) Moment of inertia J, about CF, where angle 
FCA = 0. (Principal axes as in (6).) 

7 = H 77^ h a 2 > sin 2 -f — ^ — cos 2 0. 



-fi 



B 
D C 



r 



48 ' ~ ) ' 8 

2° Hollow Circular Cylinder. — In this case 
__ ir(di 2 - d 2 2 ) j . 




Fig. 25. 



(a) Axis AB as axis of z, origin at 0. 
jl _. g _ 1r } 3(d 1 » + <y)+4P j. c _ FW + <tf) 

(6) Axis 45 as axis of z, origin at C, where OC = a. 



48 



8 



(c) Moment of inertia I, about axis CF, where angle FCA = 0. 
(Principal axes as in (6).) i 

7 = If -77T- h a 2 > sin 2 -f W - 1 — ^ — -cos 2 0. 



48 



8 



3° Thin Hollow Cylinder. — Thickness = t = 



d± — d 2 



Approximate values. In this case V = ird-Jl. 
(a) Axis AB as axis of z, origin at 0. 



A = B = w 3d s + 2l ° ; 

24 



di 2 
C =W^- 
4 



(6) Axis A B as axis of z, origin at C, where OC = a. 

d x 2 



A =B= W^^l±^l + a 2 



C = W 



(c) Moment of inertia 7, about axis CF, where angle FCA = $. 
(Principal axes as in (&).) 



W 



{ 3 dr 



2±-^ + a 2 J sin 2 + IF ^- cos 2 0. 



MOMENTS AND PRODUCTS OF INERTIA 



27 



III. Flat Homogeneous Circular Plate. 

The values for such plates will be found by making I in the 
formula for the corresponding case of the cylinder denote the 
thickness of the plate. 

Moreover, when the plate is thin, a sufficiently close approxi- 
mation will be obtained by omitting all powers of I above the 
first, in the corresponding formula for the cylinder. 

IV. Two Equal Isolated Weights Located One at e and 
one at d (Fig. 26), where Ae = Bd = c. 

W 

Let each weight = w = — > let OA = OB = z lf 



\~, 



Ad— 



and CO = a. 

(a) Axis AB as axis of z, origin at 0, axis of 

x in plane of the weights, axis of y per- 
pendicular to plane of the weights. 

A = Wzi 2 , B = W{z x 2 + c 2 ), C = Wc 2 , 

D = F = 0, E = Wczl 

(b) Axis AB as axis of z, origin at C, axis of 

x in plane of the weights, axis of y per- 
pendicular to plane of the weights. 

A = W(a? + Zi 2 ), B = W(a? + zf + c 2 ), C = Wc 2 

D = F = 0, E = Wczl 



n 
/ i "i 



_i 



Fig. 26. 



V. Homogeneous Right Circular Cone. (Fig. 27.) 

Assume that, when hollow, the outer and inner cones have 

the same half-angle at the vertex, 

and that the centers of their bases 
coincide. 

Let 

Ri = DB = radius of base of outer 

cone, 
R 2 = D'B = radius of base of inner 
cone, 
DAB = B'A'B = a, 

Ri 




h 1 = AB = 
^ = A'B = 



tana 

Ri 
tana 



Fig. 27. 



If we let G be the center of gravity, 
we have : 

For solid outer cone, 

BG =ih = jA_. 
4 4 tan a 



28 DYNAMICS OF MACHINERY 

For outer conical surface, 



3 3 tan a 

Volume of outer cone = 



irRSh 



irR2 2 h 2 



3 
Volume of inner cone = 

o 

Let CG = a. 

1° Solid Cone. — In this case R 2 = 0. 

(a) AB as axis of z, origin at A. 

A = B = !iw(*±^} C-&WV. 

(b) AB as axis of z, origin at G, the center of gravity. 

A = B = J= WRf ( 1 + 4 tan'" ), c _ 3 ^ 
80 \ tan 2 a / 10 

(c) A5as axis of z } origin at B. 

(d) AB as axis of z, origin at C, where CG = a. 

A =B = w\^Rf l \^ t J m2a + o?\> C = ^WR 1 \ 
( 80 tan 2 a ) 10 

2° Thin Hollow Cone. — Approximate solution. 

Slant height of outer conical surface = AD = Vr^ -\- hi 2 , 

Ri hi 

sin a = —7=, cos a 



VRf+hi 2 VRf + hi 2 

Let the thickness = t. .*. Ri — R 2 = = ; 

cos a hi 

We may take, for the center of gravity, that of the outer conical 
surface; hence, approximately, 

h Ri 



BG 



3 3 tan a 



Radius of elementary ring of vertical thickness dz, at distance z 
from A } is 

Ri 



MOMENTS AND PRODUCTS OF INERTIA 



29 



Volume of elementary ring at distance z from A 

2irR 1 tVR 1 2 + h 1 2 



h Y 



zdz. 



F = - (perimeter of base) (slant height) (thickness) 



(a) AB as axis of z, origin at A. 



A = B = 



irpRtf VR x 2 + h^ p» 



hr 






z 3 dz + 



2TrR 1 t\ / R 1 2 + h 1 2 C hl 



h* 



f 



z 3 dz 



= ^W(R 1 2 + 2h 1 2 ). 
2 7rpR l HVR 1 2 + h 1 2 f*i 



x 



C = *•"•""" W 1 ' ""- I z*dz = ^WR 1 2 . 

fly Jo £ 

(b) AB as axis of z, origin at G. 

A=B = ^W(9R 1 2 + 2h 1 2 ), C = \WR X *. 

(c) AB as axis of z, origin at C, where OC = a. 
A=B = ^W(9R 1 2 + 2h 2 ) + Wa 2 , C = ±WR X *. 

VI. Pulley or Flywheel. 

Assume the axis of the pulley as axis of z. 
Assume the arms to be of uniform elliptical section,* the minor 
axis of the section being parallel to OZ. 




Fig. 28. 

* Where, as is usual, the arms taper, the results given here would be 
approximately correct if the mean section is used. 



30 



DYNAMICS OF MACHINERY 



Find the moment of inertia C. 

Let r\ = outside radius of rim; 

r 2 = inside radius of rim ; 

r 3 = outside radius of hub ; 

U= radius of shaft; 

a = major semiaxis of elliptical section; 

b = minor semiaxis of elliptical section; 
W\ — weight of rim; 
W2 = total weight of all the arms; 
W3 = weight of hub ; 

d= moment of inertia of rim; 

C 2 = moment of inertia of all the arms ; 

C3 = moment of inertia of hub. 



Results : 



c 2 = Tr 2 



n 2 + r 2 2 



i 2 1 

r + o( r 2 2 -r-r 3 2 +r 2 r 3 ) 



C 3 = JT 3 



r£ + r 4 2 



Hence 



C = Ci + C 2 + C 3 = Wi 



ri 2 + r£ 



+ w- 



a 



-^ + 3 (r 2 2 + r 3 2 + r 2 r z ) 



+ W; 



n 2 + n 2 



VII. Locomotive Connecting Rod of I Section. 

For a rod in which the flanges are of uniform thickness while 
the height of the web tapers, substitute that shown in the figure, 
assuming flanges of uniform width and thickness throughout, 
and a web of uniform thickness and uniformly varying depth. 
Dimensions are indicated on the figure. 

Find the moment of inertia of the entire rod about OX. In de- 
ducing it make the following approximation in the case of the web. 

Observe that the moment of inertia of the web about OX is 
equal to the sum of its moments of inertia about OY and OZ, and 
that its moment of inertia about OY is small, hence for the mo- 
ment of inertia of the web about OX use that about OZ. We 
then obtain: 

For the web 



*-o- , ff('+¥)i 



For the two flanges, 



A ,„^ {# + V). 



MOMENTS AND PRODUCTS OF INERTIA 



31 



■-".fr— 






-HE 



=±£ 



I c 



4- 



-I __. 



-i 



t3 



^rSO- 



X | Xl 

« '<— 









1Y 



Figs. 29 and 30. 



Hence for the entire rod, 

A=A 1 + A 2 = pbL3 



1 + 



35 



+_^ (# + #). 



iz 



6* 



a 



12 V* ' & / "' 6 

Examples. — 1. Given a thin cylindrical steel basket (Fig. 31). 

Diameter = 24 inches. Depth of 
basket = 12 inches. 

Thickness of cylindrical wall = f inch. 

Thickness of bottom plate = J inch. 

The basket is attached to a vertical 
shaft CO, 12 inches long and 1 inch di- 
ameter. 

At a is attached a 4-pound weight and 
at b another 4-pound weight. 

The entire combination is supported 
at 0. 

Find the principal axes at 0, and the 
principal moments of inertia of the en- 
tire combination. 

Solution. — In making the calculation 
pounds and feet have been used as 
units. Assume the weight of one cubic inch of steel to be 0.28 
of a pound. 

Wt. = 31.68, 

Wt. = 31.68, 

Wt. = 2.64 

Wt. = 8.00 



Y/ 



Fig. 31. 



Hollow cylinder, 
Bottom plate, 
Shaft, 
Isolated weights, 



A = B = 89.76, 
A = B = 39.60 

= B = 0.88 

= 20.00, 
B = 28.00, 
D = F = 0, 



A 

A 



C = 31.68 
C - 15.84 
C = 0.00 
C = 8.00 
E = 4.00 



32 



DYNAMICS OF MACHINERY 



Entire combination, 

A = 150.24, B = 158.24, C = 55.52, D = F = 0, E = 4.00. 

Since Z) = F = 0, it follows that OF is a principal axis at O, hence 
the moment of inertia about this axis, i.e., 158.24, is one of the 
principal moments of inertia. 
To find the other two, we have 

tan 2 2 = 



A -C 
Hence we have 

«i = 2° 25' 
a 2 = 90° 
a 3 = 87° 35' 



94.72 



= 0.0845 .'. 20 2 = 4° 50' /. 2 = 2° 25'. 



71 = 92° 25', 

72 = 90°, 

73 = 2° 25'. 



ft = 90° 

02 = 0, 

03 = 90° 
Substituting these values in equation (1), we obtain 

7i = 149.97 + + 0.10 - + 0.34 - = 150.41 
h = 158.24 

7 3 = 0.27 + + 55.42 - - 0.33 - = 55.36 

Note. — In all the following examples use 0.28 of a pound for 
the weight per cubic inch of steel. 

II. Given a thin hollow steel cylinder. 




Fig. 32. 

ABCD (diameter BC = 20 inches, length = AB = 50 inches, 
and thickness = f inch), closed at each end by a thin circular disk 
(thickness = f inch) , mounted on a horizontal shaft EH (diameter 
If inches, length = 70 inches), EF = GH = 10 inches, FO = 
OG = 25 inches. Given, also, that a 1-pound weight is attached 
at A, and another at C. Find the positions of the principal axes 
of inertia at and the principal moments of inertia at O. 

„ III. Given a thin cylindrical steel basket 

(Fig. 33). 

Diameter = 24 inches. Depth of basket = 
12 inches. Thickness of cylindrical wall = 
| inch. Thickness of bottom plate = J inch. 
The basket is attached to a vertical shaft 
CO, 24 inches long and 1 inch diameter. At 
a is attached a 4-pound weight, and at b 
another 4-pound weight. The entire com- 
bination is hung at O in a ball-and-socket 
joint. Find the principal axes at O, and the principal moments 
of inertia of the entire combination. 



^ 



v/ 



*> 



ya 



Fig. 33. 



CHAPTER III. 

ACTION OF THE RECIPROCATING PARTS OF A STEAM 
OR OF A GAS ENGINE. 

Indicator Cards for Forward and Return Strokes. 

In the cylinder of a steam engine let 

Ah = head-end piston area in square inches; 
A r = area of piston rod in square inches; 
A c = crank-end piston area in square inches. 

•*• Ac = A-h A. r . 

Let us call the forward stroke that in which the piston moves 
towards the crank, and the return stroke that in which it moves 
away from the crank. Let us assume that we have a pair of simul- 
taneous cards as shown in the figures, and let us measure our pres- 
sures from the atmospheric line abb' a'. 




Let ac = eb = a'c' = e'b' = x, and let 

cd = p s = ordinate of steam line of head-end card at distance 

x from beginning of forward stroke expressed in 

pounds per square inch. 

c'd' = pj = ordinate of steam line of crank-end card at distance 

x from beginning of return stroke. 

ef = Pb = ordinate of back pressure line of head-end card at 

distance x from beginning of return stroke. 
e f f= Pb = ordinate of back pressure line of crank-end card 
at distance x from the beginning of forward stroke. 

Then we shall have, for the total force exerted by the steam 
upon the piston when it has traveled a distance x from the begin- 
ning of its forward stroke, 

F x = p 8 A h - pb f A c = A h (p s - p^ -~\ ... (1) 
33 



34 



DYNAMICS OF MACHINERY 



and for the total force exerted upon the piston by the steam when 
it has traveled a distance x from the beginning of the return stroke, 

FJ = p/A c -p b A h = A h (p/^ -p b \ ... (2) 

Hence, in order to obtain a pair of cards which represent directly 
what takes place during the forward and return strokes respectively, 
we must proceed as follows : 

1° In place of the crank-end card as drawn by the indicator, 
draw a new crank-end card, the ordinates of which, both steam and 
back pressure lines, are obtained by multiplying the ordinates of 

A 

the card, measured from the atmospheric line, by the ratio -j- c • 

2° Unite the head-end steam line with the new crank-end back 
pressure line to obtain the true forward-stroke card. 

3° Unite the new crank-end steam line with the head-end back 
pressure line to obtain the true return-stroke card. 

The result of such a process is shown in the figures, the first 
figure being the forward-stroke, and the second the return-stroke 
diagram. ' 





360° 



Fig. 35. 

Then, if we desire to know the actual force exerted upon the 
piston at any point of the forward or of the return stroke, we have 
only to multiply that portion of the corresponding ordinate of the 
forward- or return-stroke diagram that lies between the steam and 
back pressure line by A h , and the product is the required result. 

We have thus transformed the cards taken by the indicator, 
which are two separate diagrams for each end of the cylinder, 
into the two-stroke diagrams. 



ACTION OF THE RECIPROCATING PARTS IN A STEAM 

ENGINE. 

In an ordinary flywheel engine, the name " reciprocating parts" 
will be here given to the piston, piston rod, crosshead, and con- 
necting rod. 

In such an engine, while the steam pressure is exerted upon the 
piston, the resistance acts at the circumference of the drive wheel. 



RECIPROCATING PARTS OF STEAM OR GAS ENGINE 35 

Now, while the circular motion of the drive wheel, and hence that 
of the crank, is approximately uniform, the motion of the recipro- 
cating parts back and forth is variable. 

In regard to the piston, piston rod, and crosshead, the follow- 
ing facts hold, viz.: 

1° At the beginning of the stroke their velocity is zero. 

2° During the first portion of the stroke they are gaining 
velocity. 

3° At the end of the first portion of the stroke, the magnitude 
of which portion depends upon the ratio of connecting-rod length 
to crank, their velocity is a maximum. 

4° During the last portion of the stroke they are constantly 
losing velocity. 

5° At the end of the stroke the velocity becomes zero again. 

In regard to the motion of the connecting rod in the direction 
of the line of dead points, the same facts hold, except that, inas- 
much as the crosshead end of the rod has the motion of the piston 
while the crank-pin end has that of the crank pin, it follows that 
the two portions of the stroke corresponding to accelerated and 
retarded motion respectively are somewhat different from those 
for the piston. 

In the case of the connecting rod, we have also a variable motion 
at right angles to the line of dead points, which will be considered 
later. Confining ourselves, for the present, to the motion of the 
reciprocating parts in the direction of the line of dead points, we 
find that, during the first portion of the stroke, force has to be 
exerted to overcome their inertia, i.e., to impart momentum to 
them, the amount of this force becoming less and less as their 
momentum becomes greater, and finally vanishing when they 
reach their greatest momentum, somewhere near the middle of 
the stroke. After this point is passed, on the other hand, they, 
in their turn, exert force while losing velocity, the amount of force 
exerted by them becoming greater the nearer they come to the end 
of the stroke, where they lose all their momentum. 

On the other hand, it is plain that the work done during the first 
portion of the stroke in imparting momentum is just equal to that 
which they perform during the last portion of the stroke in losing 
their momentum. 

Hence, in order to ascertain what portion of the steam pressure 
exerted upon the piston at each point of the stroke is available for 
overcoming the resistance, we must deduct, from the pressure 
shown by the true stroke diagram, the amount necessary to im- 
part to the reciprocating parts the required momentum during 
the first portion of the stroke, and we must add during the last 
portion of the stroke the amount corresponding to their loss of 
momentum. Sometimes the amount of force required to impart 
the acceleration is greater than the entire pressure exerted by 



36 



DYNAMICS OF MACHINERY 



the steam at that point of the stroke, and then the flywheel is 
called upon to make up the deficiency, and drag the piston along. 
For the sake of clearness we will study this action in some simple 
case before discussing the general case. 

I. Case of Harmonic Motion (i.e., case of infinite connecting rod). 

If the motion of the piston is imparted to the crank pin by means 
of a slotted crosshead acting directly upon the crank pin, instead 
of being transmitted through a connecting rod, we have a case of 
harmonic motion, and then the motion of the piston is identical with 
the motion of the crank pin in the direction of the line of dead points. 

This kind of motion is used very frequently in steam fire engines, 
and is shown in the figure; and we will proceed to determine the 
effect of the reciprocating parts in this case. 







Fig. 36. 

Let the motion of the crank be right-handed, so that the piston 
is represented as describing its forward stroke. 

Let a = angular velocity of crank expressed in radians per second. 
t = time of describing angle cob. 
s = ed = ca = distance traveled by the piston from the 

beginning of its stroke. 
v = velocity of piston at time t. 
f = acceleration of piston at time t. 
W= weight of reciprocating parts, i.e., piston, piston rod, 

and crosshead. 
r = length of crank. 

Then we shall have 

Angle cob = at (3) 

s = r — r cos at (4) 

v = -j- = ar sin at (5) 

d 2 s 
f = -T7 2 = a 2 r cos at (6) 



RECIPROCATING PARTS OF STEAM OR GAS ENGINE 



37 



Hence, if F represents the total force required to impart this 
acceleration to the reciprocating parts whose weight is W, we shall 

have W n 

F = — a 2 r cos at (7) 

9 

Hence, the pressure per square inch of piston area (A) to be 
deducted from the pressure shown on the true stroke card is 



F W 2 

-r = — r ocr cos at. 

A gA 



(8) 




Fig. 37. 



This quantity is greatest at the beginning of the stroke, where 
it has the value 

W 

U*' (») 

and it vanishes at midstroke, where the velocity of the piston 
reaches its maximum value, which is 
ar, or equal to that of the crank 
pin. 

If the line A B represent the stroke, 
and if we lay off 

W 

AD = --rah, 

and draw through 0, the middle point 

of AB, a straight line DOC, then at any point of the stroke, as E, 

we shall have in EF the quantity 

W 

—r a 2 r cos at, 

gA 

or the quantity to be deducted from the pressure shown on the 
true stroke card at the point E. 

To illustrate this by a simple case, let us assume an engine 
taking steam, full stroke, with no compression and no back pressure, 
and let us disregard the size of the piston rod; hence call A the 
piston area on either side. Then the indicator cards, and also 
the true stroke cards, will be rectangles, whose altitudes represent 

the steam pressure in the cylinder 
above the atmosphere. 

Let AB represent the stroke, and 
let the rectangle ABEC be the for- 
ward-stroke diagram, the steam pres- 
sure (above atmosphere) behind the 
piston being AC throughout the 
stroke, and the back pressure being 
zero. 

Now compute for the given engine 
the quantity 




Fig. 38. 



W 2 

—r a 2 r, 

gA 



38 



DYNAMICS OF MACHINERY 



and lay off CF = GE equal to this quantity; then draw FG, and we 
shall have, in the trapezoid ABGF, a representation of the pressure 
upon the piston which is used in overcoming resistance through- 
out the stroke. Thus at the beginning of the stroke, while there 
is a steam pressure AC behind the piston, the portion CF is used 
up in accelerating the reciprocating parts, and hence AF is all that 
is left to overcome resistance. So, likewise, at the point H of the 
stroke, LK is used up in accelerating the reciprocating parts, and 
hence HK is all that is left to overcome resistance. At midstroke 
all the steam pressure goes towards overcoming resistance, and 
after midstroke the pressure available to overcome resistance is 
greater than the steam pressure, becoming greatest at the end of 
the stroke, when it is equal to BG. 



d' c' 





^^*?\s. 


L / 








The following figures represent the same process, carried out 
with the two stroke cards shown on page 34. While the steam 
line in the first is CDLE, this is changed when allowance is made 
for the action of the reciprocating parts to HKLM, and similarly 
for the return-stroke diagram; these new diagrams representing 
the steam pressure at each point of the stroke available for over- 
coming resistance. In all these cases we have assumed a slotted 
crosshead instead of a crank and connecting rod. 

Before giving any numerical results or numerical examples, we 
will first proceed to the general case, where we have a connecting 
rod instead of a slotted crosshead, and we shall find that the case 
of the slotted crosshead is identical with the case of a connecting 
rod of infinite length. 



II. General Case with Crank and Connecting Rod. 

When we have the case of a crank and connecting rod instead 
of a slotted crosshead, the calculations are not as simple, for 1°, 
the expression for the velocity, and also that for the acceleration 
of the piston, is much more complex than in the former case; 
and 2°, every different point of the connecting rod has a different 
velocity in the direction of the line of dead points, and hence the 
acceleration varies at each point. 



RECIPROCATING PARTS OF STEAM OR GAS ENGINE 



39 



To determine, therefore, the force necessary to impart the 
required acceleration at any point of the stroke to the piston, 
piston rod, and crosshead, we must multiply their mass by the 
acceleration of the piston. For the force needed to impart the nec- 
essary acceleration to the connecting rod, find the limit of the sum 
of the products of the mass of each small portion of the connecting 
rod by the acceleration of that portion. 

Another way of accomplishing the same result is to determine 
the portion of the weight of the rod which would be supported at 
each of its ends (crosshead end and crank-pin end), were the rod 
hung from these two points; then multiply the mass that rests at 
the crosshead end by the acceleration of the piston, and the mass 
that rests at the crank-pin end by the acceleration of the crank 
pin in the direction of the line of dead points, and add the two 
results. 

The method sometimes pursued of accounting the acceleration 
of the piston as that of all the reciprocating parts, and hence of 
multiplying the acceleration of the piston by the mass of piston, 
piston rod, crosshead, and connecting rod, is erroneous, as is also 
the method of including the mass of one-half of the connecting 
rod instead of the whole. 

We will now proceed to determine the algebraic expressions for 
the velocity, acceleration, and force required in all the different 
parts, this discussion forming the general solution of the problem. 

Let the direction of rotation be right-handed, and let 

a = angular velocity of the crank expressed in radians per 
second. 
t = time employed by the crank in passing from the dead 
point A to the position C, so that the angle AOC = at. 
I = CE = AF = BG = length of connecting rod. 
r = AO = CO = length of crank. 

s = KH = FE = space passed over by the piston, esti- 
mated from the dead point. 
vi = velocity of the piston, 
/i = acceleration of the piston. 



F E 




Fig. 40. 



Then we shall have 
5 = FE = FA + AO - OD - ED = I + r - r cos at - VCE 2 - CD 2 ; 



40 



DYNAMICS OF MACHINERY 



or, 



s = I + r — r cos at — V I 2 — r 2 sin 2 at. 

ds . , , a r 2 sin at cos at 

Vi = -j = ar sin at -\ . ; 

at VI 2 — r 2 sin 2 at 



or, 



ds 



r cos at 



vi — "T. = « r sin atf ) 1 + ■ , 

dt I VI 2 - r 2 sin 2 at 

, d 2 s „ , ( -, , r cos a£ ) 

^ 2 ( VZ 2 -r 2 sin 2 a^ 

+ ar sin a^J — ar sin a£ [7 2 — r 2 sin 2 a^]" 1 

+ r cos a£ [(Z 2 — r 2 sin 2 atf)~f (ar 2 sin at cos atf)] j 

( r (cos 2 atf — sin 2 at) r 3 sin 2 atf cos 2 at ) 

= a 2 r < cos at -\- /ln a — + 775 , . _ — 7-5 > 

{ V^ 2 - r 2 sin 2 atf (P - r 2 sin 2 at)* ) 



(1) 



(2) 



(3) 



These three equations may be put in a more convenient form 
for use, as follows: 



s = r < 1 H cos at — 

( r 




— sin 2 at 



ds 
Vi = -j- = ar sin at 



1 + 



COS atf 



, d 2 S 
*-& = <* 

Or, if we write 



cos atf-f 



- ) — sin 2 at 
r 

cos 2 atf 




sin 2 2 atf 



m 



-J — sin 2 a£ 4 



I®'- 



sin 2 a£ 



(4) 
(5) 

(6) 



A = 
B = 

C = 



cos at, » (7) 

cos 2 a£ , 

> \o) 



A 2 




— sin 2 at 



sin 2 2 atf 



Hi-' 2 



sin 2 a£ 



(9) 






we may write in place of (6) 



/ 1 = g = a 2 r(A+£ + C). 



(10) 



This gives us the acceleration of the piston, piston rod, and 
crosshead for any given crank angle at, when we know a, r, and 

(-), i.e., the angular velocity of the crank, the length of crank, and 

the ratio of connecting rod to crank. 



RECIPROCATING PARTS OF STEAM OR GAS ENGINE 



41 



If now we let W\ = combined weight of these parts, we shall 
have for the force necessary to impart the required acceleration 
to them 



„ TTi- Wid*s 
g g dt 2 



(ID 



The connecting rod will be considered later. 

ROTATIVE EFFECT. 

The steam pressure acting upon the piston is transmitted through 
the piston rod and connecting rod to the crank pin, and it is 
important, in any study of the distribution of steam, or in any 
problem of designing flywheels, to determine the pressure at right 
angles to the crank resulting from any given distribution of the 
steam pressure in the cylinder. 



>v G 




Fig. 41. 



When we know the pressure acting upon the piston at any point 
of the stroke, and the crank angle, we can determine the pressure 
upon the crank pin at right angles to the crank in one of two 
ways, as follows: 

1° Let BD = P = pressure on piston; decompose it into two 
components BE and BF, along the connecting rod and at right 
angles to the piston rod, i.e., at right angles to the guides; then is 
BE = AH the force which the connecting rod exerts upon the 
crank pin in its own direction; resolve this into two components, 
one AG at right angles to the crank, and the other AK along the 
crank; then is AG the force that balances the resistance, and this 
is called the rotative effect. 

2° Another and easier way to obtain the rotative effect ana- 
lytically is to observe that the work done by the force P per 
second is equal to the work done by AG per second, since neither 
BF nor AK do work, being merely resisted by the guides and 
the boxes respectively. Hence, if we let AG = R, we must 
have 

P X (velocity of piston) = R X (linear velocity of crank pin) . 



42 



DYNAMICS OF MACHINERY 



And substituting for these velocities their values, the former of 
which is given in equation (5), page 40, we obtain 



Par sin at 



R . , 
■p = sm at 



1 + 



1 + 



cos at 




— sin 2 at 



COS at 



= Rar. 




- ) — sin 2 at 
r. 



Hence, in order to determine the value of R when P is known 
for any point of the stroke, we need to compute tables for every 
10° (or other equal divisions of arc), and for the different ratios 
of connecting rod to crank. 

The columns that should be used for a complete table are as 
follows : 



1. 


at. 






4. 


log sin at. 


2. 


sin at. 






5. 


log cos at. 


3. 


cos at. 






6. 


sin 2 at. 


7. 


l-j — sin 2 a:£. 








8. 


-(£ 


1 — sin 2 


at\ 






9. 


log/ 


cos at 




, . 





— sin 2 o;^ 



10. 




/^ 2 
r 



11. log/l + 



1.2. log 



(' 



R 



13. 



R 



— sin 2 at 

cos at 



V 



ly 

rj 



sin 2 at , 



In the following tables we shall give only columns 1, 12, and 13 
for each ratio of connecting rod to crank. 



RECIPROCATING PARTS OF STEAM OR GAS ENGINE 



43 



TABLE I. - = 4 



TABLE II. - = 4.188. 
r 





, 1 R\ 


R 


at 


f( P ) 


P 


0° 


— 00 


0.00000 


10 


9.3353381 


0.21644 


20 


9.6259964 


0.42266 


30 


9.7846958 


0.60911 


40 


9.8850827 


0.76751 


50 


9.9501062 


0.89147 


60 


9.9898551 


0.97691 


70 


0.0096027 


1.02246 


80 


0.0123805 


1.02892 


90 


0.0000000 


1.00000 


100 


9.9734504 


0.94070 


110 


9.9329949 


0.85703 


120 


9.8780272 


0.75514 


130 


9.8066005 


0.64062 


140 


9.7143864 


0.51807 


150 


9.5920546 


0.39089 


160 


9.4172644 


0.26138 


170 


9.1167938 


0.13086 


180 


— 00 


0.00000 



at 


R 
P 


0° 


0.00000 


10 


0.21452 


20 


0.41903 


30 


0.60415 


40 


0.76179 


50 


0.88564 


60 


0.97173 


70 


1.01845 


80 


1.02680 


90 


1.00000 


100 


. 94278 


110 


. 86093 


120 


0.76033 


130 


0.64644 


140 


0.52379 


150 


0.39585 


160 


. 26501 


170 


0.13278 


180 


0.00000 



TABLE III. 



TABLE IV. 



- = 5. 

r 





. /R\ 


R 


at 


icgy 


P 


0° 


— 00 


0.00000 


10 


9.3256810 


0.21168 


20 


9.6166325 


0.41365 


30 


9.7758470 


0.59682 


40 


9.8769951 


0.75334 


50 


9.9430443 


0.87709 


60 


9.9841545 


0.96417 


70 


0.0054917 


1.01273 


80 


0.0101970 


1.02376 


90 


0.0000000 


1.00000 


100 


9.9758263 


0.94586 


110 


9.9378495 


0.86666 


120 


9.8853416 


0.76797 


130 


9.8162408 


0.65500 


140 


9.7260978 


0.53223 


150 


9.6054936 


0.40318 


160 


9.4319920 


0.27039 


170 


9.1323157 


0.13562 


180 


— 00 


0.00000 





, /R\ 


R 


at 


*•(?) 


P 


0° 


— 00 


0.00000 


10 


9.3177933 


. 20787 


20 


9.6090070 


0.40645 


30 


9.7686677 


. 58704 


40 


9.8704576 


0.74209 


50 


9.9373670 


0.86570 


60 


9.9795269 


0.95395 


70 


0.0022275 


1.00514 


80 


0.0084682 


1.01969 


90 


0.0000000 


1.00000 


100 


9.9776898 


0.94993 


110 


9.9416322 


0.87424 


120 


9.8910342 


0.77810 


130 


9.8237283 


0.66639 


140 


9.7351862 


0.54348 


150 


9.6159080 


0.41296 


160 


9.4434044 


0.27759 


170 


9.1443425 


0.13943 


180 


— 00 


0.00000 



44 



DYNAMICS OF MACHINERY 



TABLE V. l - = 5h 



TABLE VI. 



- = 6. 

r 





, / R\ 


R 


at 


log(p) 


P 


0° 


— 00 


0.00000 


10 


9.3112392 


0.20476 


20 


9.6026754 


0.40057 


30 


9.7627198 


0.57905 


40 


9.8650637 


0.73293 


50 


9.9327036 


0.85645 


60 


9.9757771 


0.94575 


70 


9.9995640 


0.99900 


80 


0.0070691 


1.01641 


90 


0.0000000 


1.00000 


100 


9.9791865 


0.95321 


110 


9.9446744 


0.88039 


120 


9.8955877 


0.78630 


130 


9.8297127 


0.67564 


140 


9.7424447 


0.55264 


150 


9.6242254 


0.42095 


160 


9.4525119 


0.28347 


170 


9.1539340 


. 14254 


180 


— 00 


0.00000 





, (R\ 


R 


at 


k-(y) 


P 


0° 


— 00 


0.00000 


10 


9.305698 


0.20216 


20 


9.597337 


0.39567 


30 


9.757711 


0.57241 


40 


9.860535 


0.72533 


50 


9.928798 


0.84878 


60 


9.972648 


0.93896 


70 


9.997357 


0.99393 


80 


0.005910 


1.01370 


90 


0.000000 


1.00000 


100 


9.980418 


0.95591 


110 


9.947166 


0.88545 


120 


9.899322 


0.79309 


130 


9.834614 


0.68330 


140 


9.748380 


0.56025 


150 


9.631023 


0.42759 


160 


9.459947 


0.28836 


170 


9.161772 


0.14513 


180 


— 00 


0.00000 



TABLE VII. 



= 7. 



TABLE VIII. - = 





, /R\ 


R 


at 


k* (p) 


P 


0° 


— 00 


0.00000 


10 


9.2968530 


0.19809 


20 


9.5888179 


0.38799 


30 


9.7497478 


0.56201 


40 


9.8533514 


0.71343 


50 


9.9226277 


0.83681 


60 


9.9677192 


0.92837 


70 


9.9938897 


0.98603 


80 


0.0041007 


1.00949 


90 


0.0000000 


1.00000 


100 


9.9823294 


0.96013 


110 


9.9510247 


0.89335 


120 


9.9050973 


0.80371 


130 


9.8421581 


0.69528 


140 


fc. 7575063 


0.57215 


150 


9 . 6414592 


0.43798 


160 


9.4713689 


0.29605 


170 


9.1737998 


0.14921 


180 


— 00 


0.00000 





, /R\ 


R 


at 


l°s( ? ) 


P 


0° 


— 00 


0.00000 


10 


9.2901002 


0.19503 


20 


9.5823264 


0.38223 


30 


9.7436932 


0.55423 


40 


9.8479058 


0.70454 


50 


9.919672 


0.82788 


60 


9.9640109 


0.92047 


70 


9.9912909 


0.98015 


80 


0.0027429 


1.00630 


90 


0.0000000 


1.00000 


100 


9.9837528 


0.96328 


110 


9.9538750 


0.89924 


120 


9.9093304 


0.81158 


130 


9.8477018 


0.70421 


140 


9.7642023 


0.58103 


150 


9.6491063 


0.44576 


160 


9.4797320 


. 33803 


170 


9.1826054 


0.15226 


180 


— 00 


0.00000 



RECIPROCATING PARTS OF STEAM OR GAS ENGINE 



45 



By means of these tables and the indicator cards we can deter- 
mine the pressure at right angles to the crank for each 10° of arc 
of crank-pin circle ; and if we lay this off from the corresponding 
point of the stroke line, and at right angles to it, we shall, by 
joining the ends of these lines, have a diagram of rotative effect. 

Of course we may apply this method to the true stroke card as 
it stands, but the results will not represent the actual available 
pressure on the crank, for a part of it is used up in accelerating the 
reciprocating parts; hence, in order to obtain the actual distribu- 
tion of pressure on the crank at different points of the stroke, we 
must use the true stroke card with the effect of the reciprocating 
parts already taken account of. 

The diagram of rotative effect resulting will represent the 
pressure on the crank at each point of the stroke, but the area of 
this diagram will not represent the work done. In order to have 
a diagram which shall show the pressure on the crank at each 
point and also the work done, we must develop the crank-pin 
circle, and, dividing this development into 10° spaces, lay off the 
pressures from these points ; then, connecting the other ends of the 
lines, we shall have the required diagram. 

The following diagrams show the result of applying this 
method to a pair of true stroke cards without taking account of 
the action of the reciprocating parts. 





360 



Fig. 42. 




Fig. 43. 



46 



DYNAMICS OF MACHINERY 



THROW OF THE CONNECTING ROD ALONG THE LINE 
OF DEAD POINTS AND ITS POINT OF APPLICATION. 

Next, as to the connecting rod, the acceleration of the crosshead 
end is given by (10), page 40, while that of the crank-pin end is 

/ 2 = ohr cos at, (12) 

as was shown under the head of harmonic motion, for this end 
moves in harmonic motion. Now, since the rod is rigid, we shall 
have the following: 




Fig. 44. 

Let AB represent the connecting rod, A being the crosshead end, 
and B the crank end. Lay off AC = /i and BD = / 2 . Then if 
we draw the straight line CD, the acceleration of any other point, 
as E, will be EF; and if we let AE = x = distance of any point 
from crosshead pin, we shall have 



EF = HE -HF =} x - GDlj 

where I = length of entire rod measured from center of crank pin 
to center of crosshead pin, or 

EF=f 1 -(f 1 -f 2 )(j) (13) 

Hence, if we let w = weight of any small particle, we shall have 
for the total force required to produce the necessary acceleration 
in the rod the expression 

*r* l wr xl 2„w f, — f„ Z n wx 

F 2 =xj[/ 1 -(/ I -/ 2 )i]-/ 1 -|-V i -V ; (13a) 

but if Xq = distance from A to the center of gravity of the rod, 
we shall have 



2 Q WX = XqZqW. 

9 ( 



h-h 



Xq 



or, 



WA, h-h 



ft-- A- 



x 



(14) 
(15) 



where W2 = entire weight of rod. 



RECIPROCATING PARTS OF STEAM OR GAS ENGINE 47 

Hence F 2 is equal to the product of the entire mass of the rod 
by the acceleration of its center of gravity, since this acceleration 

is r 

/i-(/i-/ 2 )y°' 

Moreover, we may transform (15) as follows: 

Fi = l\ fl { Wll ~^r ? ) +h ( w n)\ • • • (16) 

But if the rod were supported at A and B, the weight resting on 
these points respectively would be 



Hence, 



S A = W % 1 — j^> S b = W 2 j 



Sa r , Sb 



F,=^h + ^h (17) 

Let Xi be the distance from A to the point of application of F 2 , 
i.e., to the point of application of the resultant of the accelerating 
forces of the connecting rod in the direction of the line of dead 
points. Resolve F 2 into two parallel components acting at A and 
B respectively, and let Fa and Fb be these components; then 

F 2 = Fa -\-<Fb, 



Xi 

I 



Fb= ~r F 2 , 



Let I be the moment of inertia of the rod about the axis of the 
crosshead pin, and let p be the distance from this axis to the cor- 
responding center of percussion; then 

/ = 2 Q wx 2 = px W 2 . 

Referring to equation (15), we have 

w WA f /i-/ 2 ). 
F 2 = — Ui - —j — ■ x 1 > 

and, by taking moments about A (see (13a) ), we have 

x x F 2 = 6 z l wx - /i~/« xiwx 2 ; 

g ° gl ° 

from which we obtain 

g ( * ) 

Hence ^ ^ /i (I ~ p) + Up x 

1 fi(l — Xq) + f 2 x °* 



48 DYNAMICS OF MACHINERY 

Hence, if n = number of turns made by the crank per minute, 



= jr** j JL (P _ 2lXo + x ) + A Xo ( j _ ) 

gt 2 ( n 2 r n 2 r 

W 2 x n 2 r( /i " f 2 I 

—To — \ ~r " "" p) + ~¥" p c ' 



F 5 = 



Now the total accelerating force which acts directly at A along 
the piston rod is j? i p 

On the other hand, Fb acts at the crank pin in a direction parallel 
to the line of dead points, and its rotative effect is, consequently, 

Fb sin at. 

Hence, in order to find the total equivalent force F which, if 
applied at the crosshead A, would accelerate all the reciprocating 
parts in the line of dead points, we must add to F± + Fa a force Fz 
which would give for rotative effect Fb sin at. 

Hence, by the methods already explained, we shall have 

F = F 1 + F A + ^3 = Fi + F A + 



COS at 
1 ~r 



V r 2 



sin 2 at 



Hence, when we have drawn the true stroke diagrams from the 
indicator cards, if we wish to obtain diagrams showing the portion 
of the steam pressure used in overcoming resistance, we must 
determine the value of F for each point of the stroke, divide it by 
the area of the piston, and lay off the result vertically from the 
steam line, at the corresponding point of the stroke, downward 
when it is positive, and Upward when it is negative. 

If we apply this method to a card showing uniform pressure 
throughout the stroke, we shall not obtain a straight line cutting 
the steam line at midstroke as we did in the case of harmonic 
motion, but we shall have a curved line which cuts the steam line 
at some point other than midstroke. 

If in equation (6), page 40, we make at = and at = 180° suc- 
cessively, we shall obtain respectively, 



and 



fl*=t*r(l+j): 



- 1 + -A = - a 2 r (l - j 



These are the accelerations of the piston at the beginning of the 
forward and return strokes respectively, the first being greater, 
and the second less, than the acceleration due to centrifugal force. 



RECIPROCATING PARTS OF STEAM OR GAS ENGINE 



THROW IN A DIRECTION AT RIGHT ANGLES TO THE 
LINE OF DEAD POINTS. 

When all the preceding has been done, we have still left one 
thing out of consideration, and that is the throw of the connecting 
rod in a direction perpendicular to the line of dead points, which 
has its greatest velocity on the dead points, and loses its velocity 
entirely at the two 90° angle points of the crank. 

Hence, for a horizontal engine, it follows that the effect of the 
vertical throw is to increase the rotative effect in the upper left- 
hand quadrant (assuming right-handed rotation), to diminish it 
in the second, to increase it again in the third, and to diminish it in 
the fourth quadrant. 

As to its amount and the amount by which it alters the rotative 
effect, we have that the force of acceleration or retardation at any 
point of the stroke is equal to the mass of the entire rod multi- 
plied by the acceleration of its center of gravity in a direction per- 
pendicular to the line of dead points, and this, by a reasoning 
entirely similar to that used before, can be shown to be equal to 
the mass of the crosshead end by its acceleration plus the mass 
of the crank end by its acceleration; and since the crosshead end 
of the rod has no motion whatever, at right angles to the line of 
dead points, it follows that the throw in this direction is equal 
to the product of the mass of the crank end by the acceleration of 
the crank pin, perpendicular to the line of dead points. 




Fig. 45. 

To determine this, we proceed as follows : 

Let us use the same notation as heretofore, and we shall have 

for the motion of B in a direction at right angles to the line of 

dead points, from the dead point, 

BD = r sin at = s; 

ds 
V = -r= ar COS at. 

at 
f* = dt 2= ~ Sm ' 



50 DYNAMICS OF MACHINERY 

this being the acceleration in the direction stated, which is in this 
case a retardation. Hence, the total throw of the rod is 

— arr sin at, 

g 

its point of application being at a distance p from A. Hence the 
equivalent force at B is jl — a 2 r sin at), and its component at 
right angles to the crank is 



— a 2 r sin at cos at ) - 
9 J 



P 

l\g 

Now, in order to be able to record its effect upon the rotative 
effect diagram, we must first reduce it to pounds per square inch 
of piston area, and we shall have 



p/s b 2 . . \ ns B /47r2\ . i 

T — r a 2 r sin at cos at) = — r 7^777^ r sm at cos atf 



n. 



We therefore need to construct a table of which the columns are 
as follows: 

1. at. 

2. log sin at. 

3. log cos at. 

4. log (-) + log ( 3^Q J + log sin at + log cos at = log X. 

5. log -j- + log X + 2 log n. 

o. ii v =1 -2 '\ Qfinn / sm cos 1 7 = ro ^ a ^ lve enect 

due to vertical throw, the first four columns 
being applicable to any engine whatever. 

Deduction of the Tables for the Throw in the Direction 
of the Line of Dead Points. 

In making use of the formulae given on pages 40 and 46, it will 

be more convenient to substitute for a its value -7777-' where n = 

60 

number of turns made by the crank per minute. 

We shall thus have in place of (10), page 40, and (12), page 46, 

the following, viz. : 

fi = (n>r)(±^j\A + B + Cl, .... (20) 

K-^lm) 1 * 1 (21) 



RECIPROCATING PARTS OF STEAM OR GAS ENGINE 



51 



Now, inasmuch as we cannot determine the value of F by calcu- 
lation for every point of the stroke, we must determine it for a 
sufficiently large number of points to enable us to plot the curve. 
This may be done in two different ways, viz. : 1°, by computing its 
value for every 10° of crank angle; 2°, by computing its value for 
every tenth or twentieth part of the stroke. There is something 
to be said in favor of each method. 

If the second is pursued, it will be necessary to determine the 
crank angle for each tenth or twentieth part of the stroke (a graph- 
ical means being probably the easiest). 

In these notes, however, I shall adopt the first method. 

It is to be observed that the quantities A, B, C, A + B -J- C, 

and hence ■— and ~- depend only on the value of ( - ) > the ratio of 
n 2 r n z r \rj 

connecting rod to crank, and are independent of n and r. Conse- 
quently, they are applicable to any engine having the same ratio 

of connecting rod to crank whatever be its crank length or its 

-f 

speed. It will be noticed also that the values of -|- are the same 



n z r 



for all values of 



I 



While the intermediate columns will not be given in the tables, 
they will be enumerated here, so as to make it easy for the reader 
to deduce a table corresponding to any other ratio of connecting 
rod to crank. For this purpose he should compute and fill out the 
following columns, viz.: 



1. at. 

2. sin at. 

3. log sin at. 

4. cos at = A. 



— sin 2 at. 



12. B. 



13. log 



1 



sin 2 2 at 



n 2 



5. log COS at. 

6. log sin 2 at. 

7. log cos 2 at. 

8. log (sin 2 2 at). 




logB. 



sin 2 at 



logC. 



52 



DYNAMICS OF MACHINERY 

14. C 

15. A + B + C. 

16. log(A + £ + C). 

17. log(~Q + log(A+£ + C) 



18. 



= io <A 



A. 

n 2 r 



/4tt 2 \ 



i9 - io H3o^; +io§A=iog - 



7V-T 



20. 4- 



Of these columns only the following will be given: 1, 4, 12, 14, 15, 
17, 18, 19, 20. 



TABLE I. 

r 



at 


A 


B 


c 


A+B+C 


log (4) 


rfir 


Iog (^) 


A 

n 2 r 


0° 


1.00000 


.25000 


.00000 


1.25000 


8.1369673 


.013718.0400573 


.01097 


10 


.98481 


.23515 


.00046 


1.22042 


8.1265666 


.01338 8.0334088 


.01080 


20 


.93969 


.19221 


.00163 


1.13353 


8.0944903 


.01243 8.0130431 


.01030 


30 


.86602 


.12599 


.00300 


.99501 


8.0378847 


.01091 ,7.9775879 


.00950 


40 


.76604 


.04398 


.00394 


.81396 


7.9506604 


.00893:7.9243113 


.00840 


50 


.64279 


-.04423 


.00401 


.60257 


7.8200648 


.00662 


7.8481248 


.00705 


60 


.50000 


-.12804.00315 


.37511 


7.6142159 


.00411 


7.7390273 


.00548 


70 


.34202 


-.197021.00176 


.14676 


7.2066650 


.00161 


7.5741090 


.00375 


80 


.17365 


-.24239.00050 


- .06824 


6.8740963 


- .00075 


7.2797275 


.00190 


90 


.00000 


-.25820.00000 


- .25820 


7.4520135 


-.00283 


— 00 


.00000 


100 


-.17365 


-.24239.00050 


-.41554 


7.6586701 


-.00456 


7.2797275 


- .00190 


110 


-.34202 


-.19702!. 00176 


-.53728 


7.7702580 


-.00589 


7.5741090 


- .00375 


120 


-.50000 


-.128041.00315 


-.62489 


7.8358609 


- .00685 


7.7390273 


- .00548 


130 


• -.64279 


-.04423J.00401 


-.68301 


7.8744844 


-.00749 


7.8481248 


-.00705 


140 


-.76604 


.043981.00394 


-.71812 


7.8962543 


- .00788 


7.9243113 


- .00840 


150 


- .86602 


.12599.00300 


-.73703 


7.9075425 


-.00808 7.9775879 


- .00950 


160 


- .93969 


.19221.00163 


-.74585 


7.9127088 


-.00818 8.0130431 


-.01030 


170 


-.98481 


.23515!. 00046 


-.74920 


7.9146551 


-.00822 8.0334088 


-.01080 


180 


-1.00000 


.25000 


.00000 


-.75000 7.9151186 


-.00822 8.0400573 


- .01097 



RECIPROCATING PARTS OF STEAM OR GAS ENGINE 



53 



TABLE II. 

- = 4.188. 



at 


A 


B 


C 


A+B+C 


*(&) 


n 2 r 


lo ^) 


A 

rfir 


0° 


1.00000 


.23888 





1.23888 


8.133092 


.013586 


8.0400565 


.01097 


10 


.98481 


.22460 


.000399 


1.20981 


8.122775 


.013267 


8.0334080 


.01080 


20 


.93969 


.18355 


.001421 


1.12466 


8.091075 


.012330 


8.0130423 


.01031 


30 


.86603 


.12026 


.002609 


.98890 


8.035211 


.010845 


7.9775871 


.00950 


40 


.76604 


.04196 


.003423 


.81142 


7.949304 


.008898 


7.9243105 


.00840 


50 


.64279 


-.04218 


.003475 


.60408 


7.821154 


.006625 


7.8481240 


.00705 


60 


.50000 


-.12204 


.002726 


.38069 


7.620630 


.004175 


7.7390265 


.00548 


70 


.34202 


-.18772 


.001520 


.15582 


7.232682 


.001709 


7.5741082 


.00375 


80 


.17365 


-.23088 


.000433 


-.05680 6.794410 


-.000623 


7.2797267 


.00190 


90 


.00000 


- .25000 





-.25000 


7.437999 


-.002741 


— 00 


.00000 


100 


-.17365 


- .23088 


.000433 


-.40034 


7.642488 


-.004390 


7.2797267 


-.00190 


110 


- .34202 


-.18772 


.001520 


-.52822 


7.762857 


-.005792 


7.5741082 


-.00375 


120 


-.50000 


-.12204 


.002726 


-.61931 


7.831960 


-.006791 


7.7390265 


-.00548 


130 


-.64279 


-.04218 


.003475 


-.6815017.873525 


-.007474 


7.8481240 


-.00705 


140 


-.76604 


+.04196 


.003423 


-.72066(7.897779 


-.007903 


7.9243105 


-.00840 


150 


-.86603 


.12026 


.002609 


-.74316 7.911141 


-.008150 


7.9775871 


-.00950 


160 


-.39969 


.18355 


.001421 


-.75472 7.917845 


-.008277 


8.0130423 


-.01031 


170 


-.98481 


.22460 


.000399 


-.7598117.920764 


-.008332 


8.0334080 


-.01080 


180 


-1.00000 


.23888 


-.76112 7.921512 


-.008347 


8.0400565 


- .01097 



TABLE III. 











r 


4 1 








at 

0° 


A 


B 


c 


A+B+C 


■"»(&) 


n-r 


^ (4-) 


A 

rfir 


1.00000 


.22222 


.00000 


1.22222 


8.1272066 


.01340 


8.0400565 


.01097 


10 


.98481 


.20898 


.00032 


1.19411 


8.1171015 


.01309 8.0334080 


.01080 


20 


.93969 


.17072 


.00114 


1.11155 


8.0859862 


.01219 


8.0130423 


.01031 


30 


.86603 


.11183 


.00210 


.97996 


8.0312655 


.01075 


7.9775871 


.00950 


40 


.76604 


.03900 


.00274 


.80777 


7.9473449 


.00886 


7.9243105 


.00840 


50 


.64279 


-.03916 


.00278 


.60641 


7.8228236 


.00665 


7.8481240 


.00705 


60 


.50000 


-.11322 


.00218 


.38896 


7.6299621 


.00427 


7.7390265 


.00548 


70 


.34202 


-.17407 


.00121 


.16913 


7.2682778 


.00185 


7.5741082 


.00375 


80 


.17365 


-.21401 


.00034 


- .04001 


6.6422258 


- .00044 


7.2797267 


.00190 


90 


.00000 


-.22792 


.00000 


-.22792 


7.3978418 


- .00250 


— 00 


.00000 


100 


-.17365 


-.21401 


.00034 


-.38732 


7.6281271 


-.00425 7.2797267 


- .00190 


110 


-.34202 


-.17407 


.00121 


-.51488 


7.7517632 


-.00565 7.5741082 


-.00375 


120 


-.50000 


-.11322 


.00218 


-.61104 


7.8261268 


-.0067017.7390265 


- .00548 


130 


-.64279 


-.03916 


.00278 


-.67917 


7.8720357 


-.0074517.8481240 


-.00705 


140 


- .76604 


.03900 


.00274 


- .72430 


7.8999757 


-.00794 7.9243105 


- .00840 


150 


- .86603 


.11183 


.00210 


-.75210 


7.9163328 


-.00825 7.9775871 


- .00950 


160 


- .93969 


.17072 


.00114 


- .76783 


7.9253223 


-.00842 8.0130423 


-.01031 


170 


- .98481 


.20898 


.00032 


- .77551 


7.9296446 


- .00850 8.0334080 


-.01080 


180 


-1.00000 


.22222 


.00000 


-.77778 


7.9309130 


- .00853 


8.0400565 


- .01097 





54 



DYNAMICS OF MACHINERY 



TABLE IV. 











r 


= 5. 








at 
0° 


A 


B 


C 


A+B+C 


lo ^) 


n 2 r 


*(£) 


A 

n 2 r 

.01097 


1.00000 


.20000 


.00000 


1.20000 


8.1192377 


.01316 


8.0400565 


10 


.98481 


.19243 


.00023 


1.17747 


8.1110063 


.01291 


8.0334080 


.01080 


20 


.93969 


.15357 


.00083 


1.09409 


8.0791095 


.01200 


8.0130423 


.01031 


30 


.86603 


.10050 


.00152 


.96804 


8.0259498 


.01062 


7.9775871 


.00950 


40 


.76604 


.03502 


.00199 


.80305 


7.9447991 


.00881 


7.9243105 


.00840 


50 


.64279 


-.03514 


.00201 


.60966 


7.8251442 


.00669 


7.8481240 


.00705 


60 


.50000 


-.10153 


.00157 


.40004 


7.6421599 


.00439 


7.7390265 


.00548 


70 


.34202 


-.15599 


.00087 


.18690 


7.3116658 


.00205 


7.5741082 


.00375 


80 


.17365 


-.19170 


.00025 


-.01780 


6.2904765 


-.00020 


7.2797267 


.00190 


90 


.00000 


-.20412 


.00000 


-.20412 


7.3499421 


-.00224 


— oo 


.00000 


100 


-.17365 


-.19170 


.00025 


-.36510 


7.6024683 


-.00400 


7.2797267 


-.00190 


110 


-.34202 


-.15599 


.00087 


-.49714 


7.7365352 


-.00545 


7.5741082 


-.00375 


120 


-.50000 


-.10153 


.00157 


-.59996 


7.8181788 


- .00658 


7.7390265 


- .00548 


130 


- .64279 


-.03514 


.00201 


-.67592 


7.8699518 


-.00741 


7.8481240 


- .00705 


140 


-.76604 


.03502 


.00199 


- .72903 


7.9028019 


-.00799 


7.9243105 


- .00840 


150 


-.86603 


.10050 


.00152 


- .76300 


7.9225810 


-.00837 


7.9775871 


- .00950 


160 


-.93969 


.15357 


.00083 


-.78529 


7.9350866 


-.00861 


8.0130423 


-.01031 


170 


-.98481 


.19243 


.00023 


-.79215 


7.9388639 


-.00869 


8.0334080 


-.01080 


180 


-1.00000 


.20000 


.00000 


-.80000 


7.9431465 


-.00877 


8.0400565 


-.01097 



TABLE V. 
I 



r 



act 



o c 

10 

20 

30 

40 

50 

60 

70 

80 

90 

100 

110 

120 

130 

140 

150 

160 

170 

180 



1.00000 
.98481 
.93969 
.86603 
.76604 
.64279 
.50000 
.34202 
.17365 
.00000 
-.17365 
-.34202 

- .50000 
-.64279 
-.76604 
-.86603 
-.93969 

- .98481 
-1.00000 



.18182 
.17094 
.13955 
.09129 
.03179 
.03188 
.09206 
.14136 
.17366 
.18490 
.17366 
.14136 
.09206 
.03188 
.03179 
.09129 
.13955 
.17094 
.18182 



.00000 
.00018 
.00062 
.00114 
.00149 
.00150 
.00117 
.00065 
.00018 
.00000 
.00018 
.00065 
.00117 
.00150 
.00149 
.00114 
.00062 
.00018 
.00000 



A+B+C 



1.18182 

1.15593 

1.07986 

.95846 

.79932 

.61241 

.40911 

.20131 

.00017 

-.18490 

-.34713 

-.48273 

-.59089 

-.67317 

- .73276 
-.77360 

- .79952 
-.81369 
-.81818 



log 



\rfirj 



1126078 
1029880 
0734239 
0216305 
9427772 
8270988 
6519497 
3439218 
2705054 
3069934 
5805486 
7237608 
8115631 
8681813 
9050183 
,9285730 
,9428858 
,9505155 
,9529054 



A 

n 2 r 



.01296 
.01268 
.01184 
.01051 
.00877 
.00672 
.00449 
.00221 
.00000 
.00203 
.00381 
.00529 
.00648 
.00738 
.00804 
.00848 
.00877 
.00892 
.00897 



*(£) 



.0400565 
,0334080 
,0130423 
,9775871 
,9243105 
,8481240 
,7390265 
,5741082 
,2797267 

— 00 

,2797267 
.5741082 
.7390265 
.8481240 
.9243105 
.9775871 
.0130423 
.0334080 
.0400565 



A 

n*r 



.01097 
.01080 
.01031 
.00950 
.00840 
.00705 
.00548 
.00375 
.00190 
.00000 
.00190 
.00375 
.00548 
.00705 
.00840 
.00950 
.01031 
.01080 
.01097 



RECIPROCATING PARTS OF STEAM OR GAS ENGINES 



55 



TABLE VI. 
I 
r 



6. 



at 


A 


B 


C 


A+B+C 


Ml) 


n 2 r 


lo ^) 


A 

nV 


0° 


1.00000 


.16667 


.00000 


1.16667 


8.1070045 


.01279 


8.0400565 


.01097 


10 


.98481 


.15668 


.00014 


1.14163 


8.0975819 


.01252 


8.0334080 


.01080 


20 


.93969 


.12788 


.00048 


1.06805 


8.0686481 


.01171 


8.0130423 


.01031 


30 


.86603 


.08362 


.00088 


.95053 


8.0180223 


.01042 


7.9775871 


.00950 


40 


.76604 


.02911 


.00114 


.79629 


7.9411278 


.00873 


7.9243105 


.00840 


50 


.64279 


- .02918 


.00116 


.61477 


7.8287692 


.00674 


7.8481240 


.00705 


60 


.50000 


- .08422 


.00090 


.41668 


7.6598592 


.00457 


7.7390265 


.00548 


70 


.34202 


-.12927 


.00050 


.21325 


7.3689455 


.00234 


7.5741082 


.00375 


80 


.17365 


-.15877 


.00014 


.01502 


6.2167264 


.00016 


7.2797267 


.00190 


90 


.00000 


-.16903 


.00000 


-.16903 


7.2680717 


- .00185 


— 00 


.00000 


100 


-.17365 


-.15877 


.00014 


- .33228 


7.5615607 


-.00364 


7.2797267 


- .00190 


110 


- .34202 


-.12927 


.00050 


- .47079 


7.7128837 


-.00516 


7.5741082 


- .00375 


120 


- .50000 


- .08422 


.00090 


- .58332 


7.8059634 


- .00640 


7.7390265 


- .00548 


130 


-.64279 


- .02918 


.00116 


- .67082 


7.8666625 


- .00736 


7.8481240 


- 00705 


140 


- .76604 


.02911 


.00114 


- .73579 


7.9068104 


- .00807 


7.9243105 


- .00840 


150 


- .86603 


.08362 


.00088 


-.78153 


7.9330022 


- .00857 


7.9775871 


- .00950 


160 


-.93969 


.12788 


.00048 


-.81133 


7.9492540 


- .00890 


8.0130423 


-.01031 


170 


- .98481 


.15668 


.00014 


- .82799 


7.9580816 


- .00908 


8.0334080 


- .01080 


180 


-1.00000 


.16667 


.00000 


- .83333 


7.9608735 


- .00914 


8.0400565 


-.01097 



TABLE VII. 
1-7. 



at 
0° 


A 


B 


C 


\A+B+C 


■«(£) 


rih- 


*(&) 


A 

tfir 


1.00000 


.14286 


.00000 


1.14286 


8.0980495 


.01253 


8.0400565 


.01097 


10 


.98481 


.13428 


.00009 


1.11918 


8.0889564 


.01227 


8.0334080 


.01080 


20 


.93969 


.10957 


.00030 


1.04956 


8.0610637 


.01151 


8.0130423 


.01031 


30 


.86603 


.07161 


.00055 


.93818 


8.0123427 


.01029 


7.9775871 


.00950 


40 


.76604 


.02491 


.00072 


.79167 


7.9386007 


.00868 


7.9243105 


.00840 


50 


.64279 


- .02496 


.00072 


.61855 


7.8314313 


.00678 


7.8481240 


.00705 


60 


.50000 


- .07198 


.00056 


.42858 


7.6720884 


.00470 


7.7390265 


.00548 


70 


.34202 


-.11053 


.00031 


.23180 


7.4051699 


.00254 


7.5741082 


.00375 


80 


.17365 


-.13569 


.00009 


.03805 


6.6204112 


.00042 


7.2797267 


.00190 


90 


.00000 


-.14433 


.00000 


-.14433 


7.1994131 


-.00158 


— 00 


.00000 


100 


-.17365 


-.13569 


.00009 


-.30925 


7.5303662 


-.00339 


7.2797267 


- .00190 


110 


- .34202 


-.11053 


.00031 


- .45224 7.6954255 


-.00496 


7.5741082 


- .00375 


120 


- .50000 


-.07198 


.00056 


-.57142 


7.7970119 


- .00627 


7.7390265 


- .00548 


130 


-.64279 


- .02496 


.00072 


- .66703 


7.8642019 


- .00731 


7.8481240 


-.00705 


140 


-.76604 


.02491 


.00072 


-.74041 


7.9095288 


- .00812 


7.9243105 


- .00840 


150 


- .86602 


.07161 


.00055 


-.79386 


7.9398004 


- .00871 


7.9775871 


-.00950 


160 


-.93969 


.10957 


.00030 


-.82982|7.9590404 


-.00910 


8.0130423 


- .01031 


170 


-.98481 


.13418 


.00009 


-.85054;7.9697512 


- .00933 


8.0334080 


-.01080 


180 


-1.00000 


.14286 


.00000 


-.85714 7.9731083 


-.00940 


8.0400565 


-.01097 



56 



DYNAMICS OF MACHINERY 



TABLE VIII. 
r 



at 


A 


B 


C 


A+B+C 


i °*{i) 


A. 
rfir 


log (^) 


A 

n 2 r 


0° 


1.00000 


.12500 


.00000 


1.12500 


8.0912090 


.01234 


8.0400565 


.01097 


10 


.98481 


.11749 


.00006 


1.10236 


8.0823799 


.01209 


8.0334080 


.01080 


20 


.93969 


.09584 


.00020 


1.03573 


8.0553030 


.01136 


8.0130423 


.01031 


30 


.86603 


.06262 


.00037 


.92902 


8.0080816 


.01019 


7.9775871 


.00950 


40 


.76604 


.02178 


.00048 


.78830 


7.9367480 


.00865 


7.9243105 


.00840 


50 


.64279 


-.02181 


.00048 


.62146 


7.8334697 


.00682 


7.8481240 


.00705 


60 


.50000 


- .06287 


.00037 


.42750 


7.6709926 


.00469 


7.7390265 


.00548 


70 


.34202 


-.09867 


.00021 


.24356 


7.4266625 


.00267 


7.5741082 


.00375 


80 


.17365 


-.11836 


.00006 


.05535 


6.7831741 


.00061 


7.2797267 


.00190 


90 


.00000 


-.12599 


.00000 


-.12599 


7.1403926 


- .00138 


— 00 


.00000 


100 


-.17365 


-.11836 


.00006 


-.29195 


7.5053650 


-.00320 


7.2797267 


- .00190 


110 


-.34202 


- .09867 


.00021 


-.44048 


7.6839827 


- .00483 


7.5741082 


-.00375 


120 


-.50000 


- .06287 


.00037 


-.56250 


7.7901790 


-.00617 


7.7390265 


- .00548 


130 


-.64279 


-.02181 


.00048 


-.66412 


7.8623031 


- .00728 


7.8481240 


- .00705 


140 


-.76604 


.02178 


.00048 


-.74378 


7.9115015 


-.00816 


7.9243105 


- .00840 


150 


-.86603 


.06262 


.00037 


-.80304 


7.9447937 


-.00881 


7.9775871 


- .00950 


160 


-.93969 


.09584 


.00020 


-.84365 


7.9662188 


-.00925 


8.0130423 


-.01031 


170 


-.98481 


.11749 


.00006 


-.86725 


7.9782008 


- .00951 


8.0334080 


- .01080 


180 


-1.00000 


.12500 


.00000 


-.87500 


7.9820646 


-.00959 


8.0400565 


-.01097 



Piston and Crank Position when Acceleration in Line of Dead 

Points is Zero. 



1 

r 


4 


4* 


5 


5* 


6 


7 


8 


Per cent of stroke from ) 
middle towards H. E. ) 


5.4 


5.0 


4.6 


4.2 


3.9 


3.4 


2.9 


at corresponding 


76°. 8 


78°. 


79°. 1 


80°. 


80°. 8 


82°. 2 


83°. 1 



In applying this process to any special engine, it will be necessary 
to know — 

1° Diameter of piston. 2° Diameter of piston rod. 3° Stroke. 
4° Revolutions per minute. 5° Ratio of connecting rod to crank. 
6° Weight of piston, piston rod, and crosshead. 7° Weight of 
each end of connecting rod (Sa and Sb), whence to compute x . 
8° Distance from axis of crosshead pin to corresponding center 
of percussion. 

This last can be done by hanging up the rod on a knife edge, 
allowing it to oscillate by gravity, and counting the number of 
oscillations in a given time. 



RECIPROCATING PARTS OF STEAM OR GAS ENGINE 



57 



When these things are known, it will be most convenient to 
make out an additional table referring to that particular engine. 
The columns to be used in that table are as follows : 



1° at 



4° log 



6° log 



8 C 



Fi+Fa 



n z r 



3° ^f 



1 + 



COS at 



KOI Fb 

5 log— ■ 



— sin 2 at 



WT 



F 



B 



n 2 r / 1 + 



cos at 



IE 

y r 2 



— sin 2 at i 



B 



n 2 ril + 



cos at 



y/5- si 



sin 2 at , 



F_ 

n 2 r 



F 



9° -r, where A = area of piston, head end. 



This gives us in its last column the values of the pressures to be 
laid off from the steam line of the true stroke card. 



Numerical Examples. 

There will next be given columns 1 and 9 for a Porter-Allen 
engine, where 

Diameter of piston 10 ins. 

Diameter of piston rod 1 . 75 ins. 

Stroke 20 ins. 

Revolutions per minute 204 

Ratio of connecting rod to crank 6 

Weight of piston, piston rod, and crosshead 130 . 90 lbs. 

Weight of crosshead end of connecting rod . 55 . 66 lbs. 

Weight of crank end of connecting rod 67. 14 lbs. 

Distance from crosshead end to center of 

percussion 4 . 066 ft. 

Also the same for an Otto gas engine, where 

Diameter of piston 11 . 25 ins. 

Stroke 18 ins. 

Revolutions per minute 228 

Ratio of connecting rod to crank 4 . 188 

Weight of piston and wrist pin 164.5 lbs. 

Weight of crosshead end of connecting rod 39 . 75 lbs. 

Weight of crank end of connecting rod .... 65 . 50 lbs. 
Distance from crosshead end to center of 

percussion 2 . 71 ft. 



58 



DYNAMICS OF MACHINERY 



Also the same for a Mcintosh and Seymour Tandem Compound 
engine, where 



h. p. c. 



l. p. c. 



Diameter of piston 

Diameter of piston rod 

Stroke 

Revolutions per minute 

Ratio of connecting rod to crank 

Weight of pistons, piston rods, and crosshead 

Weight of crosshead end of connecting rod 

Weight of crank end of connecting rod 

Distance from crosshead end to center of percussion 
Head-end area of high-pressure cylinder 



11 ins. 

If ins. 
15 ins. 
240 

6 



19 ins. 

2| ins. 
15 ins. 
240 

6 



370 lbs. 
58 lbs. 
88 lbs. 
3.07 ft. 



Porter-Allen. 




F 


at 


A 




204 r.p.m. 


0° 


41.378 


10 


40.566 


20 


38.161 


30 


34.308 


40 


29.243 


50 


23 . 175 


60 


16.372 


70 


9.272 


80 


2.119 


90 


- 4.737 


100 


-11.133 


110 


-16.895 


120 


-21.917 


130 


-26.136 


140 


-29.537 


150 


-32.154 


160 


-34.013 


170 


-35.100 


180 


-35.472 



Otto Gas. 




F 


at 


A 




228 r.p.m. 


0° 


35.26 


10 


34.54 


20 


32.18 


30 


28.38 


40 


23.41 


50 


17.59 


60 


11.25 


70 


4.86 


80 


-.81 


90 


- 6.57 


100 


-11.07 


110 


-15.03 


120 


-18.10 


130 


-19.70 


140 


-20.99 


150 


-21.73 


160 


-22.15 


170 


-22.36 


180 


-22.41 



Mcintosh and Seymour Tandem 


Compound Engine. 


H. P. and L. P. Cylinders. 




F 


at 


A h ' 




240 r.p.m. 


0° 


74.19 


10 


72.62 


20 


68.20 


30 


61.13 


40 


51.72 


50 


40.55 


60 


28.64 


70 


14.76 


80 


2.75 


90 


- 9.35 


100 


-20.43 


110 


-30.18 


120 


-38.06 


130 


-44.72 


140 


-50.65 


150 


-54.69 


160 


-57.80 


170 


-59.10 


180 


-59.64 



ACCELERATING FORCES 

Also the same for an 

ALLIS TRIPLE-EXPANSION ENGINE. 



59 





H. P. C. 


I. C. L. P. C. 


Diameter of piston 


8.99 ins. 


16.01 ins. 


24.00 ins. 


Diameter of piston rod 


2.19 ins. 


2. 19 ins. 


2. 19 ins. 


Stroke 


30 ins. 

82 


30 ins. 

82 


30 ins. 

82 


Revolutions per minute 


Ratio of connecting rod to crank. 




6 
360.2 lbs. 


6 

484. 2 lbs. 


6 
774.0 lbs. 


Wt. of piston, piston rod, and crosshead. 


Wt. of crosshead end of connecting rod. . . 


100.0 lbs. 


92.3 lbs. 


100. 2 lbs. 


Wt. of crank end of connecting rod 


99.8 lbs. 


144.5 lbs. 


100.0 lbs. 


Dist. from crosshead end to center of 








percussion 


5.57 ft. 


6.25 ft. 


5.55 ft. 


H. P. Cylinder. 


I. Cylinder. 


L. P. Cylinder. 




F 




F 




F 


at 




at 




at 






A 




A 




A 


0° 


28.076 


0° 


11.330 


0° 


7.000 


10 


27.506 


10 


11 . 102 


10 


6.855 


20 


25.831 


20 


10.430 


20 


6.426 


30 


23 . 146 


30 


9.354 


30 


5.742 


40 


19.600 


40 


7.931 


40 


4.840 


50 


15.391 


50 


6.240 


50 


3.773 


60 


10.744 


60 


4.371 


60 


2.601 


70 


5.896 


70 


2.418 


70 


1.387 


80 


1.079 


80 


0.472 


80 


0.191 


90 


- 3.512 


90 


-1.387 


90 


-0.936 


100 


- 7.716 


100 


-3.097 


100 


-1.954 


110 


-11.426 


110 


-4.613 


110 


-2.839 


120 


-14.591 


120 


-5.914 


120 


-3.578 


130 


- 17 . 189 


130 


-6.989 


130 


-4.173 


140 


-19.246 


140 


-7.846 


140 


-4.631 


150 


-20.786 


150 


-8.493 


150 


-4.968 


160 


-21.854 


160 


-8.944 


160 


-5.196 


170 


-22.479 


170 


-9.209 


170 


-5.327 


180 


-22.686 


180 


-9.298 


180 


-5.370 



60 DYNAMICS OF MACHINERY 

Also for a 

FOUR-CYLINDER TRIPLE-EXPANSION CRUISER ENGINE. 





H. P. C. 


i.e. 


L. P. C. 


Diameter of piston 


36 ins. 


53 ins. 


57 ins. and 57 ins. 


Diameter of piston rod 


6f ins. 


6| ins. 


6f ins. each 


Stroke 


33 ins. 
164 
4 


33 ins. 
164 
4 


33 ins. 
164 
4 


Revolutions per minute 
Ratio of connecting rod 




to crank. 


Weight of piston, piston rod, and 








crosshead 




3068.8 lbs. 


3810.5 lbs. 


3927.3 lbs. each 


Weight of crosshead end of con- 


necting, 


y rod 




1354 lbs. 


1354 lbs. 


1354 lbs. each 


Weight of crank end of connecting 


rod 




2256 lbs. 


2256 lbs. 


2256 lbs. each 


Distance from crosshead end to 


center of npirp.nssion. 




63.04 ins. 


63.04 ins. 


63.04 ins. 








H. P. Cylinder. 


Li 


Cylinder. 


L. P. Cylinder. 




F 




F 




F 


at 




at 




at 






A 




A 




A 


0° 


90.798 


0° 


47.187 


0° 


41.519 


10 


88.896 


10 


46.189 


10 


40.638 


20 


83.317 


20 


43.246 


20 


38.043 


30 


74.360 


30 


38.527 


30 


33.883 


40 


62.521 


40 


32.297 


40 


28.393 


50 


48.462 


50 


24.914 


50 


21.888 


60 


32.952 


60 


16.797 


60 


14.739 


70 


16.847 


70 


8.395 


70 


7.343 


80 


.919 


80 


.135 


80 


.076 


90 


-14.148 


90 


- 7.622 


90 


- 6.739 


100 


-27.864 


100 


-14.618 


100 


-12.878 


110 


-40.908 


110 


-20.708 


110 


-18.213 


120 


-50.290 


120 


-25.852 


120 


-22.711 


130 


-58.917 


130 


-30.077 


130 


-26.399 


140 


-65.899 


140 


-33.448 


140 


-29.333 


150 


-71.300 


150 


-36.021 


150 


-31.568 


160 


-75.160 


160 


-37.839 


160 


-33.145 


170 


-77.480 


170 


-38.924 


170 


-34.084 


180 


-78.257 


180 


-39.286 


180 


-34.398 



ACCELERATING FORCES 



61 



Also for a 



TRIPLE-EXPANSION YACHT ENGINE. 





H. P. C. 


I. c. 


L. P. C. 


Diameter of niston 


15 ins. 

3f ins. 
30 ins. 
135 

4 1 

574 lbs. 


24 ins. 

3f ins. 
30 ins. 
135 

4 1 

^2 

759 lbs. 


37^ ins. 
3f ins. 
30 ins. 
135 

^2 

1411 lbs. 


Diamete 


t of piston rod 




Stroke. . 




Revolut 


ions ner minute 




Ratio of connecting rod to crank 


Weight of piston, piston rod, and < 
head 


3ross- 


Weight of crosshead end of connecting 
rod 


201 lbs. 
321 lbs. 


201 lbs. 
321 lbs. 


229 lbs. 
375 lbs. 


Weight of crank end of connecting rod . . . 
Distance from crosshead end to center 


of percussion 


62.14 ins. 


62.14 ins. 


65.31 ins. 


H. P. Cylinder. 


I. Cylinder. 


L. P. Cylinder. 


at 


F 
A 


at 


F 
A 


at 


F 
A 


0° 


54.834 


0° 


25.297 


0° 


17.772 


10 


53.695 


10 


24.763 


10 


17.368 


20 


50.340 


20 


23.191 


20 


16.250 


30 


44.963 


30 


20.672 


30 


14.435 


40 


37.859 


40 


17.351 


40 


12.046 


50 


29.430 


50 


13.420 


50 


9.231 


60 


20.142 


60 


9.102 


60 


6.156 


70 


10.486 


70 


4.633 


70 


2.999 


80 


0.944 


80 


0.242 


80 


- 0.071 


90 


- 8.091 


90 


- 3.884 


90 


- 2.914 


100 


-16.303 


100 


- 7.598 


100 


- 5.429 


110 


-23.513 


110 


-10.818 


110 


- 7.560 


120 


-29.640 


120 


-13.517 


120 


- 9.298 


130 


-34.694 


130 


-15.707 


130 


-10.662 


140 


-38.713 


140 


-17.420 


140 


-11.693 


150 


-41.770 


150 


-18.703 


150 


-12.438 


160 


-43.915 


160 


-19.591 


160 


-12.938 


170 


-45.189 


170 


-20.112 


170 


-13.224 


180 


-45.612 


180 


-20.285 


180 


-13.316 



62 



DYNAMICS OF MACHINERY 



Columns 1 and 6, page 50 will now be given; i.e. the throw 
at right angles to the line of dead points per unit of area of piston 
for each of the engines already referred to : 



Porter-Allen. 




204 r.p.m. 


at 


R v 


0° 


0.000 


10 


1.406 


20 


2.642 


30 


3.560 


40 


4.048 


50 


4.048 


60 


3.560 


70 


2.642 


80 


1.406 


90 


0.000 


100 


-1.406 


- 110 


-2.642 


120 


-3.560 


130 


-4.048 


140 


-4.048 


150 


-3.560 


160 


-2.642 


170 


-1.406 


180 


0.000 



Otto Gas. 


228 r.p.m. 


at 


R v 


0° 


0.000 


10 


1.293 


20 


2.430 


30 


3.274 


40 


3.723 


50 


3.723 


60 


3.274 


70 


2.430 


80 


1.293 


90 


0.000 


100 


-1.293 


110 


-2.430 


120 


-3.274 


130 


-3.723 


140 


-3.723 


150 


-3.274 


160 


-2.430 


170 


-1.293 


180 


0.000 



Mcintosh and Seymour Tandem 


Compound Engine. 




Rv 


at 


[in terms of Aft'] 




240 r.p.m. 


0° 


0.000 


10 


1.595 


20 


2.998 


30 


4.040 


40 


4.593 


50 


4.593 


60 


4.040 


70 


2.998 


80 


1.595 


90 


0.000 


100 


-1.595 


110 


-2.998 


120 


-4.040 


130 


-4.593 


140 


-4.593 


150 


-4.040 


160 


-2.998 


170 


-1.595 


180 


0.000 





ALLIS TRIPLE-E 


EXPANSION 


ENGI 


NE. 


H.I 


\ Cylinder. 


I. 


Cylinder. 


L. F 


. Cylinder. 


at 


Rv 


at 


Rv 


at 


R v 


0° 


0.0000 


0° 


.0000 


0° 


.0000 


10 


0.5723 


10 


.2932 


10 


.0802 


20 


1.0756 


20 


.5510 


20 


.1507 


30 


1.4491 


30 


.7423 


30 


.2030 


40 


1.6478 


40 


.8441 


40 


.2308 


50 


1.6478 


50 


.8441 


50 


.2308 


60 


1.4491 


60 


.7423 


60 


.2030 


70 


1.0756 


70 


.5510 


70 


.1507 


80 


0.5723 


80 


.2932 


80 


.0802 


90 


0.0000 


90 


.0000 


90 


.0000 


100 


-0.5723 


100 


-.2932 


100 


-.0802 


110 


-1.0756 


110 


-.5510 


110 


-.1507 


120 


-1.4491 


120 


-.7423 


120 


-.2030 


130 


-1.6478 


130 


-.8441 


130 


-.2308 


140 


-1.6478 


140 


-.8441 


140 


-.2308 


150 


-1.4491 


150 


-.7423 


150 


-.2030 


160 


-1.0756 


160 


-.5510 


160 


- . 1507 


170 


-0.5723 


170 


-.2932 


170 


-.0802 


180 


-0.0000 


180 


-.0000 


180 


-.0000 



PISTON AND CRANK POSITION 



63 



FOUR-CYLINDER TRIPLE-EXPANSION CRUISER ENGINE. 



H. P. Cylinder. 


I. 


Cylinder. 


L. F 


. Cylinder. 


at 


R v 


at 


R v 


at 


R v 


0° 


0.000 


0° 


0.000 


0° 


0.000 


10 


- 4.566 


10 


-2.107 


10 


-1.821 


20 


- 8.582 


20 


-3.959 


20 


-3.423 


30 


-11.562 


30 


-5.334 


30 


-4.612 


40 


-13.148 


40 


-6.066 


40 


-5.245 


50 


-13.148 


50 


-6.066 


50 


-5.245 


60 


-11.562 


60 


-5.334 


60 


-4.612 


70 


- 8.582 


70 


-3.959 


70 


-3.423 


80 


- 4.566 


80 


-2.107 


80 


-1.821 


90 


0.000 


90 


0.000 


90 


0.000 


100 


4.566 


100 


2.107 


100 


1.821 


110 


8.582 


110 


3.959 


110 


3.423 


120 


11.562 


120 


5.334 


120 


4.612 


130 


13.148 


130 


6.066 


130 


5.245 


140 


13.148 


140 


6.066 


140 


5.245 


150 


11.562 


150 


5.334 


150 


4.612 


160 


8.582 


160 


3.959 


160 


3.423 


170 


4.566 


170 


2.107 


170 


1.821 


180 


0.000 


180 


0.000 


180 


0.000 



TRIPLE-EXPANSION YACHT ENGINE. 



H. P. Cylinder. 


I. 


Cylinder. 


L. I 


. Cylinder. , 


at 


R v 


at 


R v 


at 


R v 


0° 


0.000 


0° 


0.000 


0° 


0.000 


10 


2.168 


10 


0.847 


10 


0.405 


20 


4.074 


20 


1.591 


20 


0.762 


30 


5.489 


30 


2.144 


30 


1.026 


40 


6.241 


40 


2.433 


40 


1.167 


50 


6.241 


50 


2.433 


50 


1.167 


60 


5.489 


60 


2.144 


60 


1.026 


70 


4.074 


70 


1.591 


70 


0.762 


80 


2.168 


80 


0.847 


80 


0.405 


90 


0.000 


90 


0.000 


90 


0.000 


100 


-2.168 


100 


-0.874 


100 


-0.405 


110 


-4.074 


110 


-1.591 


110 


-0.762 


120 


-5.489 


120 


-2.144 


120 


-1.026 


130 


-6.241 


130 


-2.433 


130 


-1.167 


140 


-6.241 


140 


-2.433 


140 


-1.167 


150 


-5.489 


150 


-2.144 


150 


-1.026 


160 


-4.074 


160 


-1.591 


160 


-0.762 


170 


-2.168 


170 


-0.847 


170 


-0.405 


180 


-0.000 


180 


-0.000 


180 


-0.000 



64 



DYNAMICS OF MACHINERY 



Having made out these tables, it is an easy matter to work out 
the pressure on the crank for any one of these engines from the 
indicator card, and some examples will be given here. 



I. Porter-Allen Engine 





Fig. 46. 





Fig. 47. 



- Rotative effect from True Stroke Card 

• Curve above corrected tor Horz.throw only 

■ R. E. corrected for both II. and V. throw 




Fig. 48. 



PISTON AND CRANK POSITION 



65 





II. McIntosh and Seymour Tandem Compound Engine. 

Diagram 49a shows the indicator cards from the high-pressure 
cylinder, and Fig. 496, those from the low-pressure cylinder. 



H.P.Cylinder 

Fig. 49a. 

L.P. Cylinder 

Fig. 49& 
& h 






180' •'seo" 



Fig. 49c. 



180° 



In diagram 49c the line ab is the combined steam line for the 
forward stroke, the ordinates of which at each crank angle are 
obtained from the indicator cards by first computing from them 
the total force exerted by steam on the head end of both pistons; 
this force is a certain number of pounds. The ordinate of the 
line ab corresponding to the same crank angle is obtained by divid- 
ing this force by A h ' = the area of the head end of the high-pres- 
sure cylinder. 

In a similar way are derived 

(a) the line ef, i.e. the combined steam line for the return 
stroke ; 

(6) the line cd, i.e. the combined back-pressure line for the 
forward stroke; 

(c) the line gh, i.e. the combined back-pressure line for the 
return stroke. 

For convenience in working, a table can be made out and com- 
puted for every 10 degrees of crank angle, the headings of the 
successive columns being 



66 



DYNAMICS OF MACHINERY 



High Pressure. 



1 


2 


3 


4 


Crank angle 


Ordinate from 
C. E. card 


(2) Corrected 

^1? 


Ordinate from 
H. E. card 



Low Pressure. 


Combined. 


5 


6 


7 


8 


9 


10 , 


Ordinate 

from C. E. 

card 


(5) Corrected 

hy M' 


Ordinate 

from H. E. 

card 


(7) Corrected 
, Ah 

by z? 


(3) + (6) 
C.E. 


(4) + (8) 
H. E. 



Lines Kl and mn are obtained by correcting lines ab and ef for 
the accelerations of the reciprocating parts. 

Fig. 50 is the diagram of total rotative effect, in which the 
ordinates represent the total rotative effect in pounds. 



Total Rotative Effect. 

In order to obtain a diagram which shall represent the total 
rotative effect, instead of the rotative effect per square inch of 
head end of piston, multiply each of the ordinates of the rotative- 
effect diagram already described by the area (in square inches) 
of the head end of the piston, and draw, to any convenient scale, 
a new diagram with these products for ordinates, and with the 
abscissae the same as before. 

III. Allis Triple-expansion Engine. 

In the following diagram (Fig. 51) line HH represents the result- 
ant rotative effect of the high-pressure cylinder; line II, the 
resultant rotative effect of the intermediate cylinder; and line LL, 
that of the low-pressure cylinder, while the full line represents the 
combined rotative effect in pounds. 

Weight of Reciprocating Parts. 

Whereas, in the case of a horizontal engine, the effect of gravity on 
the piston, piston rod, and crosshead have no effect upon the throw 
in a direction at right angles to the line of dead points, and whereas 
what effect is had by the weight of the connecting rod is usually 
neglected, it becomes necessary in the case of a vertical engine to 
correct the rotative-effect diagram for the rotative effect due to 
the weight of the reciprocating parts. 



PISTON AND CRANK POSITION 



67 







c 

t— 




o 


•w *°. P. - § P. 
< o i-I Oco ci 


1 1 1 1 


i l 


N' 


Moment 
Scale 

T 










r 




o 

O 








CO 






























/ 








3 


/ 






,3 


{ 




Oi 


\ 




















CO 













t 

m 






CO 

1— 1 


;ative E 
Scale 

3000 IbT 
2000 IbsT 

iooo ibiT 


A i 

o o 
o o 
o o> 


1 ^ 




o 








a 






1-1 






































































CIS 

|9 














3 






» 


















CO 


DV. 














































u. 














o 



o# 



'°*r- 



-i — t- 

I / 



Vir 



I y v 



w-v 



/I \ 



10 



^ x 



68 



DYNAMICS OF MACHINERY 



To do this, take the weight of piston, piston rod, and crosshead, 
and add in the weight of the crosshead end of the rod; divide the 
sum by the area of the piston in square inches, and call this W. 
Find the rotative effect of W, for every 10 degrees of crank angle, 
and lay it off. 



R.E. For AUis Intermediate Cylinder [horz.] 
. R.E. for same cylinder regarded as vertical 




In Fig. 52, the full line shows, in the case of a certain engine, 
the rotative-effect diagram uncorrected for gravity, while the 
dotted line shows the rotative-effect diagram corrected for gravity. 



FLYWHEELS. 

The flywheel is a wheel with a heavy rim, having consequently 
a large moment of inertia, whose function is to store up energy 
during the acceleration of the machine or motor due to change of 
energy or load, and to give it up during retardation; and, in con- 
sequence of its large moment of inertia, to keep the fluctuation 
of speed within certain small limits. 

Fluctuation of speed may be due to change in the energy supplied 
or in the resistance, or in both, and the moment of inertia of the 
flywheel must be sufficient to take care of both. 

If there are a number of places where fluctuations occur, it may 
be best to place a separate flywheel at each; but if one is used to 
control them all, it should be placed near where the fluctuations 
are largest. Before we can set out to determine the proper size 
of flywheel to use for any particular machine, we must first deter- 
mine the value of the greatest fluctuation of energy for that 
machine, which we will call 

AE. 

For punching and slotting machines, this is nearly the whole 
energy expended during one operation. 

In the case of a steam engine, one portion of AE is due to the 
variation of the load, while another portion is due to variation of 
energy in the engine itself. We can find it by drawing our dia- 
gram of rotative effect on the development of the crank-pin circle, 
determine its area, and then draw a straight line, making a rec- 
tangle equal in area. 

Then measure the areas of the portions of the diagram that lie 
respectively above and below this line, and reduce to work units, 
and the largest of them is AE. Thus (in Fig. 50) if OPURVSWTIO' 
be the diagram, draw OA of such length that OAPRSTAO'O shall 



FLYWHEELS 69 

equal the area of the figure; then will AE be represented by the 
largest of the four following areas, viz.: RPUR, RVS, SWT, and 
TIO'A' + AOP. In this case the largest is RPUR, and this rep- 
resents the amount of work that is alternately stored up in and 
removed from the flywheel, in consequence of the fluctuation of 
energy in the engine itself. In this case, i.e, that of the Mcintosh 
and Seymour engine, AE is 2305 foot-pounds, while W, the work 
performed per revolution, is 10,406 foot-pounds. Hence the ratio 

AE 

— = 22 

W 

In proportioning a flywheel, however, for any given engine, 
we cannot generally obtain a card, as the engine is not yet built. 
Hence, we should draw a theortical card, i.e., one in which the cut-off 
is that where the greatest fluctuation of energy will probably 
occur (as a rule, this is at the longest cut-off with which we should 
be liable to operate the engine) or at the greatest cut-off at which it 
is desired to preserve the given regulation; then for the expansion 
line draw an hyperbola, proceed to obtain the diagram of rotative 
effect, and also AE; then we have the fluctuation of energy due to 
the action of the engine itself, and now the problem may assume 
various forms: 

1° We may desire to know the dimensions of flywheel necessary 
to preserve this regulation with this AE only. 

2° We may desire to know, with a given flywheel, what regu- 
lation we may expect to maintain with this fluctuation of energy 
in the engine, assuming the load constant. 

3° We may have the means of ascertaining the amount of 
load that may be thrown on or off; and this being added to that 
arising in the engine, we may wish to proportion our flywheel so 
as to preserve the desired regulation under the two sources of 
variation combined. 

4° We may, not knowing our variation of load, as the engine 
builder cannot know it, simply allow a certain proportion of the AE 
of the engine for the greatest possible fluctuation of load. In some 
cases 50 per cent is allowed and sometimes more. When the value 
of AE, which the flywheel is to take care of, is known, and also the 
limits of variation of speed required, to determine the proper moment 
of inertia of the wheel. For this purpose we proceed as follows: 

Let / be the moment of inertia of the wheel about its axis. 

Let a = mean angular velocity of the wheel in radians per second. 

Let — th of the mean speed be the greatest allowable fluctuation 
m 

of speed. 

Then we have 

Greatest angular velocity = a 1 1 + » — 1 • 

Least angular velocity = a ll — - — ) • 



70 DYNAMICS OF MACHINERY 

/ J. 1 



2m 



Greatest actual energy of wheel = ^— a 2 ( 1 + 

If 1 V 

Least actual energy of wheel = p— a 2 1 — - — • 

2g \ 2 m) 

Hence, since the difference of these two is the greatest fluctua- 
tion of energy, we have 

AE =J-ao 2 \(l +^-V-fl L 



2 g (\ 2 m) \ 2 m 

.:AE=fa4*) = 1 ^ (D 

2 g \m) mg 

From this we easily obtain 

I=^AE (2) 

When we know the values of AE and ao, and also the limits of 
fluctuation of speed that we wish to preserve, we can easily find 
the moment of inertia that the wheel must have in order to control 
the speed within the required limits. 

The next thing to be done is to design the wheel so that it may 
have this moment of inertia. 

We may, if we choose, make a rough approximation by neglect- 
ing the effect of the arms and hub and considering only the rim of 
the wheel. 

If we do this, and if we let 

W = weight of rim, 

T\ = outside radius of rim, 

r 2 = inside radius of rim, 



we shall have 



I = Wrl+l£l (3) 



Hence, combining this with (2), we have 

W= 2mgAE ■ (4) 

but if we wish to work more accurately we must take into account 
the moment of inertia of the arms and of the hub. (See VI, 
page 29.) 

Acceleration of the Flywheel when the Load is Suddenly Changed. 

Suppose the steam pressure, the load under which the engine 
is running, the events of the stroke (the cut-off being, say, 0.5), 
and consequently the speed to be constant. 



FLYWHEELS 71 

Let ao = this constant speed in radians per second. 

No= the number of revolutions per minute, corresponding 
to ao. 

Inen a = — — • 

Let Wq = work per minute in inch-pounds. 

M = corresponding driving moment in inch-pounds, 

ZttJS q 

M r = moment corresponding to longest cut-off at which the 
engine ever runs. 
Y] r = angular distance moved through by the governor 
weight when the engine passes from no load to full 
load. 

Now suppose that the load Wq is suddenly decreased to one 
corresponding to a shorter cut-off (as 0.3). 

Let a\ = speed under the new load in radians per second. 

Ni = number of revolutions per minute corresponding to a\. 

TW TNl 

Inen on = -^r • 

Let Wi = work per minute in inch-pounds under new load. 
Mi = driving moment corresponding to load Wi. 
-q — angular motion of governor weight while engine passes 
from driving moment M to whatever the driving 
moment becomes at end of time t. 

Now Mi is constant. But the driving moment, which is M at 
the instant when the change occurs, will decrease as the cut-off 
shortens. It is sometimes assumed as an approximation that at 

end of time t it becomes M — M r — • In that case the unbalanced 

moment acting to accelerate the flywheel would be 

M - Mi -M r ^- 

But the last term is small, and when we are only concerned with the 
determination of the speed up to and at the next cut-off, the last 
term may be neglected, and the unbalanced moment causing accel- 
eration of the flywheel is M — Mi. Hence if we let 

a — angular velocity in radians per second t seconds after the 
change, where t is less than the time to the next cut-off, 
N = number of revolutions per minute corresponding to a, 
I = moment of inertia of flywheel, units being pounds and 

inches, 
g = 386 inches per second, we shall have 



72 DYNAMICS OP MACHINERY 

Ida . da (Mp ^ MQg 

f (Afo - M 1 )g . (Mp-MQg 

This equation enables us to determine the velocity of the flywheel 
at any time t after the change of load, provided t is less than or 
equal to the time when the steam distribution is changed, by 
change of cut-off, or change of compression. 

Were the load suddenly increased, instead of being decreased, 
there would be a retardation instead of an acceleration. 

Example. — Let N = 350 .'. a = 36.652 radians per second. 

I = 142,800 (units pounds and inches), M = 6123 inch-pounds. 
Mi= 7894 inch-pounds; then M Q - M 1 = - 1771 inch-pounds. 

Hence -=- = ._ or ^^ — = — 4.787 radians per second. 

dt 142,800 

= 36.652 - 4.787 1. 

If we assume that the steam distribution does not change till the 

30 3 

next cut-off, then we have, time of one stroke = 7^. = tt= second. 

oou 00 
3 

Hence at end of first stroke we have t = 5= second, and hence 

00 

= 36.652 - (4.787) — = 36.242 radians per second. 

Ar (30) (36.242) OAa no 
.'. N= - — — = 346.08 r.p.m. 

7T 



McIntosh and Seymour Engine. 
Velocity and Displacement Curves. 

Let M = Rr = moment of rotative effect in inch-pounds. 
6 = angular velocity in radians per second. 
/ = moment of inertia of flywheel about its axis. 



Then we have 

dB g 



de -V M - g f 

dt-I M > 6 -lJ 



Mdt. 



The values of M can be obtained from the rotative-effect diagram. 
The scale to which the ordinates were laid off was 1 inch = 20 
pounds per square inch of piston area, the piston area is 95.03 
square inches, and the crank length 7J inches. Therefore this 
curve will represent moments of rotative effect, i.e., values of M> 
to a scale of 1 inch = (20) (95.03) (7.5) = 14,255 inch-pounds. 



FLYWHEELS 73 

To obtain the velocity curve, lay off as ordinates the successive 
sums of the mean ordinates of the moment curve (Fig. 50), these 
ordinates being measured from the line A A, and we thus obtain 
the velocity curves AFDEA, Fig. 53. To ascertain its scale 
proceed as follows : 

i" = 1,870,000 pounds (inches) 2 , g = 386 inches per second, 
.*. j = 0.0002064. Moreover, if At be the time corresponding to 

10 degrees crank angle, since the r.p.m. = 240, At = —^r second. 
Hence the scale of the ordinates of the velocity curve is such that 
1 inch = (14,255) (0.0002064) U^) = 0.0204 radians per second. 

Now 

Let m = mean velocity in radians per second, so that 

0i = 6 — B m represents the angular velocity from the mean. 
Let y) = angular displacement. 
Then we have 



x] = f 0i dt. 



Following a process similar to that described above, using the 
velocity curve, we obtain the displacement or r\ curve BJKLMB, 
in which 

0.0204 
1 inch = * . = 0.0001417 radians displacement. 

Hence, when these curves have been constructed and their scales 
have been determined, we can find the angular amount in radians 
by which any point on the circumference of the wheel is ahead of 
or behind the position it would have occupied had the speed not 
varied at all. 

In this case the greatest displacement is at crank angle 67.5° 
and measures on the diagram 16.75 inches. This is equivalent to 
0.00237 radians = 0.136° = 0.0724 of an inch at the circumference 
of the flywheel. 



Centrifugal Tension in the Rim of the Pulley. 

The following is an approximate method of computing the 
centrifugal tension in the rim of a pulley, i.e., the rim tension due 
to centrifugal force, which would exist in the case of a pulley with 
a thin solid rim, were the stresses in the rim unaffected by the 
arms. 

It is sometimes erroneously given as the method for computing 
the actual stresses in the rim. 



74 



DYNAMICS OF MACHINERY 

Displacement Scale 



>— 
OS 

r 


h- h- M 

>*>: tO © 00 OS *». t© OtO 

r p i 5 p p p p * p 


»*\ OS 00 © *o »K» OS 00 

r p i 5 i 5 p r p p 











era 

Or 
CO 




Velocity Scale 



FLYWHEELS 75 

Let R = mean radius of rim in inches. 

A = area of cross section of rim in square inches 

w = weight of material in pounds per cubic inch. 

g = acceleration due to gravity = 386 inches per second. 

V = linear speed of rim in inches per second. 

V 

a = r- = angular velocity in radians per second. 

F = total centrifugal tension in rim in pounds. 

p = centrifugal force per inch length of rim in pounds. 

F . 

t = -j = centrifugal tension per square inch in pounds. 

Then we have 

wAV 2 

Hence we have the case of a thin hollow cylinder subjected to an 
internal normal pressure of p pounds per inch of circumference. 
Hence 

F = pR = , 

9 

_wV 2 

9 

Example. — Let the pulley be of cast iron, for which w = 0.2604, 
and let the speed of the rim = 1 mile per minute; then V = 
1056 inches per second. We then have 

. (0.2604) (1056) 2 _ 

t = ^^ = 752 pounds per square inch. 

If / = allowable working tensile strength per square inch of 
the material and Vo = allowable rim speed, and were this method 
correct for the determination of the actual stress per square inch 
in the rim, we would have 

w^V - . Tr Igf 



V W 



9 
Example. — Given / = 1000 pounds per square inch, find Vo. 

Vo = i/- — ~ nn^A — = 1217 inches per second = 6085 feet per 
V 0.2604 

minute. 

Stresses in the Rim and Rim Joints of Pulleys Due to Centrifugal 

Force. 

The two cases that need to be considered are 

1° When the pulley is cast in one piece; 

2° When it is cast in sections united by bolts. 



76 DYNAMICS OF MACHINERY 

The first case does not include the largest wheels, for, it being 
impracticable to cast them whole, they are cast in sections and 
bolted together. 

The considerations, however, that affect the pulleys cast in one 
piece affect also those made in sections, though other stresses 
also come into play. 

Beginning, therefore, with the first case, viz., pulleys cast in one 
piece, observe that, were there no force exerted by the arms on 
the rim, the only stress in the rim would be the centrifugal 
tension. 

The stresses actually existing in this first case are 

1° A direct tensile stress which is a portion only of the centrif- 
ugal tension; 

2° Stresses due to the bending of the portion of the rim between 
two adjacent arms. 

By way of further explanation, we may observe that, in order to 
have no bending, the arms would have to stretch, in consequence 
of their own centrifugal force, enough to allow the rim to stretch 
to the amount called for by the centrifugal tension. 

When the stretch of the arms is less than this, the rim is con- 
fined at the points of junction with the arms, and hence arises 
bending. 

A discussion of the case of solid pulleys is given by Professor 
Unwin in his Machine Design, the last portion of which introduces 
some approximations. 

We will proceed with what seems to the author to be a simpler 
demonstration of the first part, and then continue with a fuller 
discussion. 




Fig. 54. 

Unwin' s notation will be used. 

Let adbc represent the portion of the rim between two consecu- 
tive arms oa and oc. 

Let a = angle aob = angle hoc = one-half the angle between two 
consecutive arms. 
= variable angle aod = twice angle aoe, so that aoe = eod. 



FLYWHEELS 77 

R = oa = distance from center of hub to center of rim in 
feet. 

v = linear velocity (in feet) of center of rim per second. 
A — area (in square feet) of cross section of rim. 
G = weight of the metal in pounds per cubic foot. 

g = 32.16 feet per second. 

F = pull exerted by each arm on the rim, so that the shear- 

F 

ing force in the rim close to the arm = — • 

S = Shearing force in rim at variable point d, where angle 
aod = 4>. 

T\ = direct tension in rim in tangential direction just over the 
arm. 

T = direct tension in rim in tangential direction at variable 
point d, where angle aod = 4>. 

M = bending moment in rim in foot-pounds at variable point 
d, angle aod = </>. 

M\ = bending moment in rim in foot-pounds at its junction 
with the arms. 

k = F + l-v 2 . 
3 g 

7*1 = distance (in feet) from center of hub to outer end of arm. 
r 2 = radius of hub in feet. 
I = moment of inertia of cross section of rim about neutral 

axis, units being pounds and feet. 
y 2 = distance from neutral axis of rim to outside. 
2/i = distance from neutral axis of rim to inside. 

a — stress at outside of rim due to bending only (in pounds 

per square foot). 
(7i = stress at inside of rim due to bending only (in pounds per 

square foot). 
p 2 = stress (in pounds per square foot) at outside of rim. 
Pi = stress (in pounds per square foot) at inside of rim. 
Ei = modulus of elasticity of the metal (in pounds per square 

foot). 
E = modulus of elasticity of the metal (in pounds per square 

inch) . 

A i = area of cross section of arm in square feet, when the 

arm is of uniform section throughout. 
AR = elongation of arm due to the action of centrifugal force 

(in feet). 

F 

Si = =- = shearing force just next to the arm. 



73 DYNAMICS OF MACHINERY 

■S 2 = = shearing force halfway between two consecutive 

arms. 
T 2 = direct tension halfway between two consecutive arms. 
M 2 = bending moment halfway between two consecutive arms. 
Consider the forces acting on a portion ad of the rim, when the 

angle aod = 4> (variable). 

The forces are the following, viz. : 

1° The centrifugal force acting on this part of the rim, the 
resultant of which acts along the line oe outwards and 
equals 

|-4„) (chord ad) = 2-Av 2 sin- 0. 
\g RJ g 2 

2° The direct tension T h acting at a in a direction tangent to 

the arc ad towards the right. 
3° The direct tension T (variable) acting at d, in a direction 

tangent to the arc ad towards the left. 

F 

4° The shearing force — acting just to the left of a in the direc- 
tion ao. 

5° The shearing force S (variable) acting at d in the direction 
do. 

6° A bending moment Mi at a. 

7° A bending moment M (variable) at d. 

Resolving forces along the directions oe and ad and imposing 
the conditions of equilibrium, we have (+ upwards and + to the 
left) 

C 1 1 1 1 F 1 

2—Av 2 sin^cf) — Tsin-(/) — Tisin-</> — S cos-4> — t;Cos-</> = 0. (1) 

Q —i Zj A Zi Zi lj 

1 -i -| ip -| 

Tcos- $ — jTiCos- </> — *Ssin- <j> + — sin- <j> = 0. . (2) 

Z Z, Z Zi z 

M = M 1 -^sin</)-2ri^sin 2 i0 + f2-JLz; 2 sin^0Vi?siniA (3) 

In (3) the signs are so chosen that the bending moment is posi- 
tive when the bending tends to make the rim concave outwards. 
When = 2 a, either (1) or (3) gives 

G F 
T l = -Av 2 -^ t cota (4) 

9 2 

Substituting this value of 7\ in (1) and (2) and solving for S 
and T, we obtain 

s = F sin (a - 0) _. 

2 sin a ' 

T = G Av2 _F co S ( a -4>) 

g 2 sin a 



FLYWHEELS 79 

and (3) becomes 

, , ,.. . FR ( cos (a — d>)) ,„ N 

M = M 1 -\-- Fr ] cot a ^ — [ .... (7) 

2 ( sin a ) 

To find Mi observe that when = a, the slope is zero. 
Hence / Md<f> = 0; consequently, substituting the value of M 
from (7), integrating, and solving for Mi, we have 

Mi = F r(l~ cota )> - (8) 

and substituting in (7), we have 

M = FR ( 1 _ cos (a - 0) > 

2 } a sin a y 

Equations (5), (6), and (9) give the values of the shearing force, 
direct tension, and bending moment respectively, at the variable 
point d, where aod = </>. 

On the other hand, when <f> = or <j> = 2 a, we have 

^ F „ G A _ F ,-. ,Pi? / 1 \ 

Si = 77; 7\= -Az; 2 - -cota;ilfi = -^- cot a . 

Z g Z I \<x I 

Moreover, when </> = a, 

G F 

S 2 = 0;T 2 = -Av 2 --coseca;M 2 = 

g 2 

These equations are all identical with those given by Professor 
Unwin. They give the shearing force, direct tension, and bending 
moment, at any point, in terms of F, the force exerted by each arm 
on the rim. 

Hence it becomes necessary to find the value of F, so as to 
substitute it in the above equations. To do this in the case of 
arms of which the section varies, would lead to more or less com- 
plexity, but it should be done whenever necessary; the only case 
considered here, however, will be that of arms of uniform section 
throughout; and the results may sometimes be applied with 
tolerable accuracy, when the variation is small, to those whose 
average section is the same as that of the uniform arm considered. 

Let Ci = centrifugal force of the portion of the arm between 
the rim and the end of a variable radius p, then we shall have 

n Gv * a T 1 j Gv2 a n 2 - p 2 , in . 

and the total stretch of the arm due to the entire force acting 
upon it is 

.0 G v 2 A fW-p 2 dp , frFdp 




gB? l J n 2 AtE^Jr, A 1 E 



80 

This reduces to 



DYNAMICS OF MACHINERY 



AB = (r, - r0 \\% 1 e{^ - \™ - l^+j^E \- (11) 

Moreover, the total stretch of the portion abc of the rim is that 
due to the tension T, and hence we have, disregarding the slight 
change of shape of the arc, 



2aAR = 



hence 



-x 



"TRj R {„ G 



AM 



, ,2a- Av 1 - F 
AE I g 






AR =^ 



RiG 2 

= — { - v 2 — 
E (g 



F 



2A 



a 



(12) 



Hence by equating (11) and (12), solving for F, and reducing, 



F=--v 2 

3g 



3- 



r ± 



r 2 \ 2 fri + J r 2 



— r„\2 



R 



R 



1 r 1 — r 2 



(13) 



2A 



or if we write F ={- 






3 - 



A; = 



fl 



fl 



1 n — r 2 



(14) 



Ai R ' 2Aa 

As a summary of the equations deduced we have the following, 
viz.: 

F sin (a — 0) > 



<£ = </> 



sin a 



[ )= - 



= 



£ =- 

r-^,2 F cosja- <ft) . 

# 2 sin a 

M= FR/1 cos (a- 0)y 

2 \« sin a 

Si = - g ; • • • ■ 

T x =-^ 2 -^-cota:; 

2 

Mi= -n~[ cot a 



<j) = a 



(15) 
(16) 
(17) 
(18) 
(19) 

(20) 

(21) 

Tz = -Av i -^coseca; (22) 

Q 2 



)' 



S 1 -= 0; 
G 

n* FR I 1 

Mi = -7T- cosec 



2 \a 



■)■■ 



(23) 



FLYWHEELS 81 

and if we write 



-fif* 



we shall obtain 



♦-o;^-T + -r-^) 1+ «li«-U + -r) oota ji ; (24) 

, . T 2 M 2 y 2 G A W Ryi./Ryi 1\ "R,o^ 

*-a;p,- T — r -,-^i+^_-X+^_lL-_j COB eo«j;(2B) 

k= ) R l\ f I (26) 

1 r\ — r 2 , 1 

It may be of interest, in any special case, to compute the values 
of the direct tension per square inch, and of the stress due to bend- 
ing separately. If this is desired, the following are the formulas 
to be used: 



T! G . (• k >. M lVl G 9 kR yi (l , \ 

— - =— if ; 1 — — — cot a> , ai = T = — V- - -7-1 cot a); 

A g I 6A ^ ii g 6 J \a /' 

T 2 G A^ k ) . M 2 y 2 G 2 kRy 2 f 1\ 

4 ^ ( 64 J i 2 £ 6 J \ «/ 

As an example, let us take a pulley (shown in the figure) 48 inches 
outside diameter and 12 J inches face, with a rim -ft- inch thick, 
and a rib 1 inch square in the mid- 
dle of the inside of the rim; the (WMmM^^^M^^MMJ^s ^ 
number of arms of the pulley be- |§i 

ing six, each arm being elliptical ^ 

in section, the major diameter Fig. 55. 

being 2f inches, and the minor 
diameter l\ inches; the diameter of the hub being 7 J inches. 

k 
We shall find ~ = 0.813; and for v = 88 feet per second, i.e., 
A 
a rim speed of 1 mile per minute, the greatest value of p occurs 
when <j> = 0; and then pi = 5657 pounds per square inch. 

Stresses in the Rim and Rim Joints of Bolted Flywheels. 

Sometimes this bolting is done halfway between two consec- 
utive arms, and sometimes over the arms. In the latter case, 
however, the amount by which the rim of the wheel projects 
beyond the arms in a direction parallel to the shaft is often so 
great that the outer portion receives little or no reenforcement 
from the connection of the rim with the arm. 



82 



DYNAMICS OF MACHINERY 




Fig. 56. 



In both cases the joint is almost invariably the weakest part 
of the structure. 

Proceeding now to our discus- 
sion, take first the case when the 
joint is halfway between two con- 
secutive arms, and use the same 
notation that was employed in 
the earlier part of this discussion. 
We should first make the follow- 
ing calculation, which disregards 
whatever effect there may be 
due to the overhang of the rim 
beyond the arm in a direction 
parallel to the shaft. 

In Fig. 56 let ad = distance 
of center of gravity of the rim 
section from the outside of the rim; let ebf be the line of the axis 
of the bolt or bolts, and c the lowest point where the flanges come 
in contact. 

The stresses in the bolts, rim, and flanges are different according 
as one or the other of the two following conditions holds, or a con- 
dition intermediate between the two, the extremes occurring when 
1° The bolts are set up very tightly, and when the rim and 
flanges are very stiff; 

2° The bolts are so loose that the two parts of the joint do not 
touch each other. 

Beginning with the first case (see Fig. 56), we have for the forces 
acting at the joint the tension T 2 (applied at a point so near a 
that it will be practically near enough to consider it at a), together 
with the bending moment M 2 ; but this combination is equivalent 

to a single force T 2 applied at a h where aai= -=-i and is laid off 

i 2 

outwards from a. 

Now, inasmuch as the fastenings are not in line with the single 
resultant force, T 2 acting at a h sl bending moment arises in the 
joint, which in this case is taken up by the bolts and flanges and 
not by the rim, and we consequently have, if S is the total stress 
in the bolts, that n r 

S=T 2 °^- 

DC 

Besides this, the greatest fiber stress in the flanges should be 
determined from the bending moment they have to bear, but this 
is so simple a proceeding that I shall not stop to deduce a formula. 

Taking up now the second case, when the bolts are so loose that 
the two parts of the joint do not touch each other, we find that the 
entire discussion of the stresses that act in a solid pulley no longer 
finds any application here; for there can be no bending moment 
M 2 at the joint. 



FLYWHEELS 83 

Hence, in this case, the resultant force acting at the joint is F 
(the centrifugal tension), applied so near a that we can consider it 
at a. 

Then, since the bolts are loose, the total stress in the bolts is 
only F , but the bending moment F (ab) is taken up by the rim. 

In the actual case the stresses may be either of those described 
above, or anywhere intermediate between them, and are liable to 
vary in their distribution according to the speed and the conse- 
quent amount of yielding of the different parts. 

After having made the calculations described above, which, as 
stated, disregard the effect of the overhang of the rim beyond the 
arms, we should, when the overhang is at all considerable, carry 
out a similar set of calculations, substituting F (the centrifugal 
tension) for T 2 , and the point of application a for a h thus determin- 
ing what would be the stresses near the edge of the rim if the over- 
hang is so much that this is not reenforced by its connection with 
the arms. 

Then if (as would probably be true in most cases when the joint 
is between two consecutive arms) the stresses determined by the 
former set of calculations are greater than those determined by the 
latter, we should design the wheel so that it will resist the former 
stresses with safety; but if, as might happen, the stresses, or some 
of them, came out greater in the latter set of calculations, the 
wheel should be designed so as to bear with safety the greatest 
to whichever set they belong. 

We will now proceed to consider the case where the rim joints 
are directly over the arms, which is the most usual case in large 
built-up fly-band wheels. 

If we were to make our calculations by disregarding the effect 
of the overhang of the rim beyond the arms in a direction parallel 
to the shaft, i.e., to determine the stresses that would arise if the 
overhang were very small, we should find that the tension Ti at 
a, together with the bending moment M h would be equivalent to 
a single resultant tension Ti at a point a h which would now be 

Mi 
below instead of above a, and where aa±= -=- ; i.e., the resultant 

1 1 

tension would be T h and its point of application ai would be 

below a. 

As long as this point ai remained above b, the mode of calcu- 
lation outlined in the other case would apply; while if the point 
a i were to go below b (not a usual case), the tendency to pivot 
would be around d instead of around c. 

The first would be the case in wheels with a very small overhang, 
and also would apply to the portion of the rim directly over the 
arms in those with a considerable overhang, except that the vari- 
ous modes of fastening the rim to the arm would come in to mod- 
ify the calculations; and it would be useless to attempt here any 
detailed discussion of these various modes of attaching the rim to 



84 DYNAMICS OF MACHINERY 

the arm, as they all differ in detail; and the calculations for deter- 
mining the stresses in one would not be suitable for another 
arrangement. 

Next consider the case of the outer edge of the overhang. Unless 
the flanges or lugs are so stiff that their deflection is so slight as 
not to allow the outer edge of the rim to increase in diameter to 
the extent necessary to correspond to the action of the centrifugal 
tension (with the effect of the arms absent), the outer edge of the 
rim will be in the same condition that it would be if there were no 
arms; and the mode of calculation to be followed will be explained 
even at the risk of seeming repetition, because this is one of the 
most frequently occurring cases. 

The total hoop tension in the rim will be F (the centrifugal 
tension), applied at a point so near a (see Fig. 56) that it may 
practically be considered as applied at a. 

Now, inasmuch as the fastenings are not in line with the force 
F , a bending moment arises, and two cases are conceivable: 

First, that the bending is taken up by the fastenings, i.e., the 
bolts and flanges, and not by the rim; 

Second, that the bending is taken up by the rim and not by the 
bolts. 

In the first case, which would occur when the bolts are set up 
tightly, we should have 

In this case there is a bending moment in the flange at b equal 
to F (ab), but there is no bending moment in the rim. 

In the second case, which would occur if the two parts of the 
rim did not touch each other, the stress in the bolts is only Fo, 
but the bending moment F (ab) is taken up by the rim. 

The pulley should be so designed that the bolts and flanges are 
strong enough to resist the stresses if they occur, as described in 
the first case, and that the rim and flanges are strong enough to 
resist the stresses if they occur as described in the second case. 

The following problem will serve to illustrate the above dis- 
cussion. Assuming the rim shown in Fig. 56, the bolts being 
eight inches apart on centers, the total centrifugal tension to be 
resisted by one bolt, neglecting the flange (which in an actual 
case should be considered), at a rim speed of one mile per minute, 
is: 

n WAv* (450) (8 X|) (88) 2 anQn , 

Case I. Fq = = 00/ , — - = 9030 pounds, 

g 386 

S = F (g) = 9030 3 >< 5 - 5 = 16,555 pounds. 

The area of the bolt at the root of the thread is 1.30 square inches. 

16 555 

' = 12,734 pounds per square inch = stress in bolt. 
l.o 



FLYWHEELS 85 

Stress due to bending in flange, 

My 9030X2.5X12X1 KO - ft 
a= ~T = 6.5 X 2 X 2 X 2 = 521 ° P0unds - 

Case II. S = F = 9030 pounds, 

9030 - _ , , , ■ .. 

.. » = 5947 pounds per square men = stress in bolt. 

Stress in rim due to bending, 

My 9030 X 2.5 X .75 X 12 _ OK , 

* = ~T = 8 X 1.5 X 1.5 X 1.5 = 7525 P ° Unds - 

Direct stress in rim due to centrifugal tension (per square inch) 

= 752.5 pounds. 
Total stress per square inch = 8277.5 pounds. 
Stress due to bending in flange, 

My 9030 X 1.75 X 1 X 12 on __ , 

a = T = 8X2X2X2 = 2963 p0Unds ' 

Stresses in the Arms of a Pulley Due to Centrifugal Force Only. 

The total direct tension in any one arm (if straight) at a distance 
p from the axis of the pulley is T = F + Ci, where the value of 
C\ is given in equation (10), page 79, and that of F is given in 
equation (13), page 80. 

The greatest value of C\ occurs when p = r 2 , and is 

_ G v 2 . ri 2 — r 2 2 
Cl ~gR Al 2 

Hence if t\ = direct tensile stress in pounds per square inch in the 
arm, at the hub, we have 

F_ G ^ ri 2 - r 2 2 
h ~A l + gR 2 

Besides this tensile stress, there is also a bending moment to be 
borne by the arms when the velocity of the wheel changes. The 
greatest value of this total bending moment would be equal to the 
entire turning moment transmitted from the shaft to the wheel, 
and if the wheel is a flywheel, and not a drive wheel, this total 
bending moment would be equally divided among the arms. 
Hence 

Let Mi = total turning moment transmitted from the shaft 
to the wheel. 
n = number of arms. 
M = bending moment borne by one arm. 

.*. M= — - 
n 



86 DYNAMICS OF MACHINERY 

H.P. = horse power transmitted. 

N = number of revolutions per minute. 

We then have 

(12) (33,000) H.P. . , 
Mi = ~ — ' ,/ inch-pounds, 

ZirJN 



and 



M = — - 
n 



Let Ii = moment of inertia of section of arm about neutral axis.. 
y = distance from neutral axis to outside fiber of arm. 
t 2 = outside fiber stress due to bending moment. 
t = greatest stress in arm. 

Then we have 

and hence 

* * _i_ * F .G v 2 r! 2 -r 2 *My 

t = tl + t2 = i + ^R^^ + 7;' 

In the cases where the flywheel is also a drive wheel, the extra 
stresses due to the driving needs to be considered. It is quite 
common to consider the total bending moment due to this cause 
equally divided between the arms. 

Experiment has shown, however, that the arm nearest the point 
where the belt leaves the pulley on the tight side has to bear a 
larger bending moment than the others. Indeed, the stresses in 
arms and rim due to the action of the belt vary with the thickness 
of the rim. 



ROTATIVE EFFECT IN GAS ENGINES. 

Rotative-Effect Diagram for Gas Engines. 

CASE I. 

Fig. 57 is an indicator card for the firing cycle of a four-cycle 
single-acting single-cylinder llj-inch-by- 18-inch Otto gas engine, 
running at 228 revolutions per minute, and regulated on the hit- 
or-miss principle. The suction-curve is taken at atmospheric 
pressure throughout its length. Fig. 58 is a diagrammatic view of 
the same engine, the direction of rotation being indicated by the 
arrow. 

The crank angle is assumed to be zero, when the piston is at the 
head-end dead point, at the beginning of the suction stroke. In 
Fig. 59 the line Aa(3y8B represents, to the same scale, the same 



ROTATIVE EFFECT IN GAS ENGINES 



87 



card, changed to a continuous card, for one complete firing cycle, 
which is, of course, divided as follows, viz. : From 0° to 180° suc- 
tion, from 180° to 360° compres- 
sion, from 360° to 540° expan- 
sion, and from 540° to 720° 
exhaust. 

On the other hand, in order 
to obtain from this continuous 
card the rotative-effect diagram, 
for one firing cycle, we proceed 
as follows, viz.: 





Fig. 57. 



Fig. 58. 



1° Reverse the ordinates of the compression, and of the exhaust 
lines, as they represent negative work, and thus obtain 
the lines A a, afc, f3y8, and 5iB. 

2° Correct these lines for the action of the reciprocating parts, 
and thus obtain the lines A 2 a 2 , a s (3 2 , fisyzh, and 8sB 2 . 

3° Multiply each of the ordinates of these lines by the ratio of 
rotative effect corresponding to its crank angle, and thus 
obtain the ordinates for the rotative-effect diagram, Fig. 60. 

If now the engine fires every second revolution, i.e., if there are 
no misses, then this diagram; Fig. 60, as it stands, gives us the 
following information, viz. : 

1° The mean value AC = BD of the rotative effect, determined 
by dividing the resultant area of the rotative-effect dia- 
gram by the length AB (i.e., the length corresponding to 
two revolutions). 

2° The value of AE for use in designing the flywheel. 

3° The angular-velocity diagram of the flywheel can be plotted. 
The angular-displacement curve of the flywheel can be plotted. 
The rotative-effect curve, if the scale be suitably changed, 
will represent the moment of the rotative effect corre- 
sponding to each crank angle. 
If the ordinates are measured from CD instead of from AB, 
they will represent the values of the respective moments 
of the rotative effect above and below the mean. 

7° We can also compute the ratio of AE to the average work 
done in one revolution. In this particular case this ratio 
is 2.1. 

If, however, there are miss cycles, as fire 1 miss 1, or fire 7 miss 1, 
then the diagram should be so extended that it will include one 
complete cycle of operations. 



4° 

5° 



6 C 



88 



DYNAMICS OF MACHINERY 



3 
re 

» 360 

o 




ROTATIVE EFFECT IN GAS ENGINES 



89 



. Evidently the more the misses, 
for the same firing cycle, the smaller 
the mean rotative effect AC — BD, 
the smaller the average work per- 
formed in one revolution (which 
will be called W), and the greater 

^ + - AE • 

the ratio ^77- • 

W 

In the case of the given card, 



with fire 1 miss 1, -777 = 

W 



6.4. 



This, however, is a light load, 
hence this value of the ratio would 
not be suitable to use in designing 
the flywheel. 

In the case of the given card, 

•? - » - ■■ £ - -■ 

This is a normal load, and this 
value of the ratio would be suitable 
to use in designing the flywheel. 

Diagrams for Other Styles 
of Engines. 

In order to illustrate the general 
character of the rotative-effect dia- 
gram in other styles of gas engines, 
the same indicator card has been 
assumed, and the assumptions 
made regarding the action of the 
reciprocating parts are explained 
in each case. 

CASE 11. 

Single-acting two-cylinder twin 
engine. Fig. 61 is a diagrammatic 

view of 
such an 
engine. 
The suc- 
tions fol- 
low each 
other 
Fig. 61. every 

360°, and hence also the compres- 
sions, expansions, and exhausts. 
Since the cylinders are alike, the 





90 



DYNAMICS OF MACHINERY 



assumption has been made that the reciprocating parts of each 
cylinder are alike. 

Fig. 62 is the rotative-effect diagram which may be obtained by 
either of the two following methods, but which was obtained by 
method I. 

Method I. — Construct for each cylinder a rotative-effect dia- 
gram like that shown in Fig. 60. Draw these two diagrams 360° 
apart, and then combine them. 

Method II. — Construct for each cylinder a continuous card 
with the ordinates of the compression and exhaust reversed, as 
shown in Fig. 59, lines Aa, cfti, fiy8, and 8iB. Draw these two 
cards 360° apart and then combine them. Then correct them for 
twice the effect of the reciprocating parts of one cylinder. Then 
multiply each ordinate of the resulting curve by the ratio of rota- 
tive effect corresponding to its crank angle, and with the products 

AE 
as ordinates plot the rotative-effect diagram. r===- would probably 



W 



be about one-half that for case I. 



CASE III. 



Single-acting two-cylinder tandem engine. Fig. 63 is a dia- 
grammatic view of such an engine. The cycle is the same as in 
case II, i.e., the suctions follow each other every 360°. 




Fig. 63. 

To construct the rotative-effect diagram, proceed as follows, 
viz.: Correct the cards of the rear cylinder for the difference in 
areas of the pistons. Construct for each cylinder a continuous 
card with the ordinates of the compression and exhaust reversed. 
Draw these two cards 360° apart, combine them, and correct the 
result for the effect of the entire reciprocating parts of the engine, 
in the manner already explained in the case of the Mcintosh 
and Seymour tandem steam engine. Then multiply each ordi- 
nate of the resulting curve by the ratio of rotative effect corre- 
sponding to its crank angle, and with the products as ordinates plot 
the rotative-effect diagram. The diagram for this case will not be 

AE 
given here, -^r would probably be about the same as for case II. 



W 









L 




r 




Fig. 64. 
the complete cycle being as follows: 



CASE IV. 

Double-acting one-cylinder 
engine. Fig. 64 is a diagram- 
matic view of such an engine. 
The explosions occur here 180° 
apart every two revolutions, 



ROTATIVE EFFECT IN GAS ENGINES 



91 






a 

era 

OS 
os 




a 

era 
OS 




92 



DYNAMICS OF MACHINERY 





H. E. 


C. E. 


0°-180° 

180 -360 

360 -540 

540 -720 


Suction 

Compression 

Expansion 

Exhaust 


Compression 
Expansion 

Exhaust 
Suction 



To obtain the rotative-effect diagram, proceed as follows, viz. : 

1° Correct the crank-end cards for the difference in areas of the 
two sides of the piston, and plot the H. E. and C. E. cards 
in the relative positions shown in Fig. 65, or by the table 
above. 

2° Combine these cards, and then correct the result for the 
effect of the reciprocating parts. The result is shown in 
the full line in Fig. 65. 

From this deduce the rotative-effect diagram in the usual way. 

A W 

The result is shown in Fig. 66. In this case ^rr — 1.3. 



W 



CASE V. 




Fig. 67. 

Double-acting two-cylinder tandem engine. Fig. 67 is a dia- 
grammatic view. The expansion strokes occur every 180° apart, 
the complete cycle being: 





H. E. Cylinder. 


C. E. Cylinder. 


H. E. 


C. E. 


H. E. 


C. E. 


0°-180°... 
180 -360. . . . 
360 -540.... 
540 -720.... 


Suction 
Compression 
Expansion 
Exhaust 


Exhaust 
Suction 
Compression 
Expansion 


Expansion 
Exhaust 
Suction 
Compression 


Compression 
Expansion 
Exhaust 
Suction 



The curve should be corrected and combined as already explained 

AE 
in the case of the tandem steam engine. -^ would probably be 

about the same as for case VI. 



CASE VI. 



Twin double-acting two-cylinder engine. Fig. 68 is a dia- 
grammatic view. The cycle is the same as for case V. As the 
explosions occur every 180°, we only need to displace 180°, the 



ROTATIVE EFFECT IN GAS ENGINES 



93 



rotative-effect diagrams for two double-acting single-cylinder 
engines. 

1 




Fig. 68. 
The full line in Fig. 69 shows the result. In this case 

^ = 0.2. 
W 



CASE VII. 




Fig. 70. 

Single-acting two-cylinder opposed engine. Fig. 70 is a diagram- 
matic view. The cycle is the same as for case IV. 

In this case, the piston displacements (measured from the 
head ends) for the two cylinders, for the same stroke, do not corre- 
spond. However, in the rotative-effect diagram, the results will 
be correct if we combine the ordinates for the same crank angle, 
and then correct for the effect of the reciprocating parts by adding 
the 10° value to the 170°, the 20° to the 160°, etc. The rotative- 
effect diagram from these results will be similar to the full line in 

AE 
Fig. 66. -jjTT would doubtless be somewhat larger than for case II. 

CASE VIII. 




Fig. 71. 

Twin single-acting four-cylinder opposed engine. Fig. 71 is 

a diagrammatic view. The cycle is the same as for case V. 

Combine the rotative-effect diagrams for two two-cylinder opposed 

engines, displaced 180°, as the explosion curves follow each other 

AE 
in this case. The results will be similar to Fig. 69. -==r would 

W 

be small, probably about the same as for case VI. 



94 



DYNAMICS OF MACHINERY 



QP3 

Oi 
CO 




ROTATIVE EFFECT IN GAS ENGINES 



95 



CASE IX. 



Twin double-acting tandem four-cylinder engine. Fig. 72 is 
a diagrammatic view. This is a combination of two double-acting 



i n 



j i 



i 



*-. — ^o ^ 



I 




Fig. 72. 

two-cylinder tandem engines, and the rotative-effect diagram is 
to be obtained by combining the two separate ones in the same 

AE 
way as in the case of a steam engine. •=- would be small, depend- 
ing on the angle between the cranks. 

Action of Reciprocating Parts. 

Case when the Path of the Crosshead Pin Does not Pass Through 
the Center of the Crank-pin Circle. 

Let ACB be the crank-pin circle, and FG the path of the cross- 
head pin, F and G being the dead points. Let OH = a. Then 

we have 

OF = l + r, OG = l-r, FH = V(l -f-r) 2 - a 2 , 

GH =V(l-r) 2 -a 2 . 




Fig. 73. 

Draw A' OB' parallel to the path of the crosshead pin, and let 

CO A i = at, ( 

/. EH = r cos at + V> - (a + r sin at) 2 



= r cos at -f- r 




l^ 2 

r 



- (; + sin at ) 



FE = FH - EH = s = V(l + r) 2 - a 2 - r cos at 

— r 



vf-OH' 



96 DYNAMICS OF MACHINERY 

Hence we have 



ds 



sin at + 



I- + sin at) 



cos at ( - + sin at 




g+ sin at) 



4- d 2 s 



cos at + 



cos 2 a:£ — sin at( — \- sin at) 
r/ - g. + sin «*)' 




+ 



(f + ^in a<j 



cos 2 atf( — |- dn atf 



®H?wr 



At F we have s = 0, and v = 0, and from the geometry of the 
figure a 



sin atf = — 



l -+i 

r 



Hence, making this substitution in the value of /, we obtain at F 

/i =a 2 r 




At G we have s = V(l + r) 2 - a 2 - V(7 - r) 2 - a 2 , v = 0, and 
from the geometry of the figure 

sinatfi = 



I-i 

r 



Making this substitution in the value of f h we obtain at G, 
/ 1= -o: 2 r — 



*-7 




ACCELERATION OF RECIPROCATING PARTS 97 

Rotative Effect. 

If we denote by R the rotative effect, and by P the corresponding 
pressure on the piston, we shall have 

cos atl- -\- sin at 

4. X V 

sin at 



Rar = Par 



Hence 



\/© 2 - (5 



R . ,, 

p = sin at + 



-f- sin at 
cos atl - + sin an 



Equivalent Pressure on Piston. 

All the formulae, discussions, and conclusions on pages 47 and 
48 apply equally to the present case, provided /i represents the 
acceleration of the crosshead pin, i.e., the value of /i deduced for 
this case. 

The value of F, however, will no longer be that deduced on 
page 48, but will be as follows, viz. : 

F = F! + F A + ^ — . 

cot an- + sin ail 
1 + 



Throw in a Direction at Right Angles to the Path of the 

Crosshead Pin. 

The discussion and the equations deduced on pages 49 and 50 
for the ordinary case will all apply to this case, except that in 
lines 5 and 6 from the bottom of page 49 the words " at right angles 
to the line of dead points, from the dead point," should read "at 
right angles to the line AiB h from the time when the crank coin- 
cides with OA x." And to the discussion on pages 47 and 48 
should be added the following: 

Observe that when the crosshead pin is at F, the beginning of 

the stroke sin at =— j—, — , hence the crank pin is at a vertical 

I + r 

ar 
distance below the line AiBi equal to =—. — . Moreover, when the 

I + r 

crosshead pin is at G, the end of the stroke sin at — , _ , hence 

the crank pin is at a vertical distance above the line AiBi equal 



98 



DYNAMICS OF MACHINERY 



to 



ar 



I — r' 



Hence, while the crosshead pin travels from F to G, 



the angle described by the crank is t — sin -1 



a 



sin -1 7— — and 

I — r I + r 

while the crosshead pin travels from G to F, the angle described 

by the crank is r + sin -1 ■= sin -1 7— 

I — r I + r 

Moreover, the value of / 3 , the vertical acceleration, becomes at F 



a l r 



a 



Z + r' 



and at G, — a 2 r 



a 



l-r 



Graphical Methods. 

While it is the belief of the author that, for constant use, the 
analytical methods already described are the easiest and the most 
accurate, nevertheless, for the benefit of those who prefer graphical 
methods, some of them will be given, for the determination of 

(a) The components, parallel and perpendicular to the line of 

dead points, of the velocity of any point of the rod. 

(b) The components, parallel and perpendicular to the line of 

dead points, of the acceleration of any point of the rod. 

(c) The respective throws of the entire rod, in directions parallel 

and perpendicular to the line of dead points. 

(d) The resultant throw of the entire rod in magnitude and 

direction. 

(e) The point of application of the throw parallel to the line of 

dead points. 
(/) The point of application of the throw at right angles to the 

line of dead points. 
(g) The point of application of the resultant throw. 

Graphical Construction for the Components of the Velocity of Any 

Point on the Rod. 

Proposition I. — Let Fig. 74, 0, be the center of the crank 
shaft, OC the crank in length and position, and CW the connecting 
rod in length and position. 




Fig. 74. 



GRAPHICAL CONSTRUCTION 99 

If we make the construction shown in Fig. 74, that is, prolong 
WC till it meets at A, the perpendicular drawn to OW at 0, draw 
CR perpendicular to OW, and RN equal to OR, and at right angles 
to OW, and join C with A', WA' being equal to OA, and N with W 
by straight lines, then, when the angular velocity of the crank in 
circular measure is unity, we shall have: 

1° OC will represent, in magnitude, the linear velocity of C (the 
crank pin), its direction being perpendicular to OC; CR 
will represent the component of the linear velocity of C 
parallel to the path of the crosshead pin OW, and OR 
will represent the component of the linear velocity of C 
in a direction perpendicular to OW. The above is so 
evident from the construction of the figure that it is un- 
necessary to demonstrate it. 

2° OA will represent, in magnitude, the velocity of W (the 
crosshead pin), the direction of this velocity being, of 
course, along OW, and hence at right angles to OA. 

3° If we assume any point H on the rod, and draw through it 
the line LM at right angles to OW, then will GL repre- 
sent the component of the velocity of H in a direction 
parallel to OW, and GM will represent the compon- 
ent of the velocity of H in a direction at right angles to 
OW. 

Demonstration of 2° and 3°. — Draw CO' perpendicular to OC 
and equal to it in length; then will CO' represent the velocity of C 
in magnitude and direction. Resolve CO' into two components, 
CP and PO' respectively, in the direction of and at right angles to 
WC; then since CP represents the component of the velocity of 
C along WC, it also represents the component of the velocity of 
W along WC. Now lay off WB equal to CP, and draw BD at 
right angles to WC; then will WD represent the velocity of W in 
the direction in which it moves, i.e., along OW. 

But if we draw Oa perpendicular to CA, it is easy to prove that 
the triangle COa is equal in all its parts to the triangle O'CP, and 
that the triangle aOA is equal in all its parts to the triangle BWD; 
hence it follows that OA = WD; and hence follows 2°. 

Now LG is parallel to CR, and the point G divides WR in the 
same ratio as the point H divides WC, and since WA' is laid off 
from W parallel to RC, and WA' is equal in magnitude to the 
velocity (along OW) of W, and RC is the component along OW of 
the velocity of C, it follows that GL is the component in a direc- 
tion parallel to OW of the velocity of H ; and, similarly, it may be 
shown that GM is the component of the velocity of H at right 
angles to OW. Hence follows 3°. 

Proposition II. — If the angular velocity of the crank, expressed 
in radians per second, is a instead of unity, then, if we lay off OC 



100 



DYNAMICS OF MACHINERY 



in Fig. 74 to represent ar, the linear velocity of C, instead of mak- 
ing it equal to r, and if we lay off CW equal to 

'I 



-I M = cd, 

then we shall have that OA will represent the linear velocity of 
W, that GL will represent the component of the velocity of H 
parallel to OW, and that GM will represent the component of the 
velocity of H at right angles to OW. 

Demonstration. — Changing the angular velocity of the crank 
from unity to some other value of a, while retaining the same 
ratio of connecting rod to crank, results in multiplying the veloc- 
ities of all points in the system by a. Hence, if OC be laid off 
equal to ar instead of r, then Fig. 74 will give directly the velocities 
of the different points. Hence follows the truth of the proposition. 

Graphical Construction for the Components of the Acceleration of 

Any Point in the Rod. 

While there are several different graphical constructions for the 
acceleration of the crosshead pin, and hence for the accelerations 
of the other points in the rod, the general method pursued in nearly 
all of them may be described as follows (see Fig. 75) : 




P„ i\ d 2 



Fig. 75. 



Let the angular velocity of the crank, in radians per second, be 
unity. Let, as before, OC represent the crank, and CW the con- 
necting rod. Find a point S' on CW, such that CS' X CW = AC 2 ; 
then from S' draw a perpendicular S'S to CW, meeting OW at S. 
Then we have: 

1° OC will represent in magnitude and direction the acceleration 
of C; also, if CR be drawn perpendicular to OW, meeting 
OW at R, OR = Cc will represent the component of the 
acceleration of C in a direction parallel to OW, and CR = 
Oc will represent the component of the acceleration of C 
at right angles to the line OW. The above is so evident 
from the construction of the figure that it is unnecessary 
to demonstrate it. 



GRAPHICAL CONSTRUCTION 101 

2° OS will represent the acceleration of W, of course, in the 
direction OW. 

3° If from any point P on the rod we draw Pppi parallel to 
OW to meet CS in p, and from p we draw pp 2 perpendicu- 
lar to OW, then will pip represent the component of the 
acceleration of P in a direction parallel to OW; pp 2 will 
represent the component of the acceleration of P at right 
angles to OW, and Op will represent the resultant accel- 
eration of P in magnitude and direction. Indeed, the 
line CS is called the acceleration image of the rod. 

Demonstration of 2° and 3°. — Resolve the acceleration of C, 
or OC, into components along and at right angles to the rod WC; 
then will Ca represent the former and Oa the latter. But when, 
as is the case here, the angular velocity of the crank is unity, OC 
will also represent the linear velocity of C, and OA that of W; and 
as each is at right angles to the velocity which it represents, there- 
fore will the third side CA of the triangle OCA represent the veloc- 
ity of C relatively to W, or, in other words, the velocity of C when 
W is considered as fixed, or the velocity with which C revolves 
about W; and, of course, CA is perpendicular to the velocity which 
it represents. 

CA 2 

Now, since CS f = yuj7 } anc ^ smce @A represents the linear veloc- 
ity of C relatively to IT as a center, therefore CS r represents 
the acceleration of C in the direction CW relative to IF as a 
center. 

We thus have that Ca represents the component of the actual 
or absolute acceleration of C, and that CS' represents the accel- 
eration of C relatively to W; hence, since Ca and CS' have opposite 
directions, their sum, and not their difference, or Ca + CS' = S'a, 
will represent the component of the acceleration of W in the direc- 
tion WC. Hence the actual acceleration of W is represented, in 
magnitude and direction, by OS. Hence follows 2°. 

Draw Ppp and Cc, both parallel to OW; then the point p\ divides 
the line Oc in the same ratio as P divides WC. Now, since OS 
is laid off from the end corresponding to W (along OW), and 
Cc = OR is laid off from the end corresponding to C equal to the 
component of the acceleration of C parallel to OW, therefore 
Pip = Op 2 will represent the component of the acceleration of P 
in a direction parallel to OW. 

Again, the point p 2 divides SR in the same ratio that p divides 
SC, and hence in the same ratio that P divides WC. Then, since 
RC represents the component of the acceleration of C in a direc- 
tion at right angles to OW, and since W has no acceleration in a 
direction at right angles to OW, and as p 2 p is parallel to RC, it 
follows that p 2 p represents the component of the acceleration of P 
in a direction at right angles to the line OW. Hence follows 3°. 



102 DYNAMICS OF MACHINERY 

Proposition IV. — If the angular velocity of the crank is a 
instead of unity, then if we lay off OC, in Fig. 75, to represent a 2 r, 
the acceleration of the crank pin in a radial direction, and if we 

lay off CW equal to f-]a 2 r = a 2 l, then will OS represent the accel- 
eration (along OW) of W, while Op 2 represents the component 
parallel to OW of the acceleration of P, and p 2 p represents the 
component at right angles to the line OW of the acceleration of P. 

Proposition V. — If, instead of accelerations, we wish to obtain, 
from Fig. 75, pressures on the piston required to produce the 
accelerations, we only need to make OC represent the product 
of the mass concentrated at C and the acceleration at C, then will 
OC, CR, and OR represent forces required to produce the accel- 
erations. In other words, if the weight concentrated at C is W c , 
then we must lay off OC to represent 

W c , lirWcWr 
— a 2 r = > 

Q 9 

and construct Fig. 75. 

Proposition VI. — If, on the other hand, we wish to obtain, 
from Fig. 75, pressures per square inch of piston area required to 
produce the accelerations, we only need to make OC represent 
the product of the mass concentrated at C and the acceleration 
at C, divided by the area of the piston, then will OC, CR, and OR 
represent forces per square inch of piston area required to produce 
the accelerations. In other words, if the weight concentrated at 
C is W c , then we must lay off OC to represent 

W c , 4:>irW c N 2 r 

— r a 2 r = -j > 

gA gA 

and construct Fig. 75. 

Determination of the Point S'. 

Three different graphical solutions will be given for the deter- 
mination of the point S' on the rod, so that CS' X CW = (AC) 2 . 




Fig. 76. 

The first is shown in Fig. 76 and consists in drawing AB parallel 
to OW until it meets in B the line OC (produced if necessary), and 



GRAPHICAL CONSTRUCTION 



103 



then drawing from B the line BS f at right angles to OW until it 

meets CW in S'. This is the method given by Mohr. 

Demonstration. — 

CS f AC 
From the similar triangles CS'B and CAO we have -^ = -^c • 



From the similar triangles ACB and OCW we have 



CB CO 
CB CO 



AC CW , 
By multiplication j£ = ^ ' •'• CS' X CW = (AC) 2 . Q. E. D. 




Fig. 77. 

The second method is shown in Fig. 77 and consists in describ- 
ing a circle on CW as diameter, and another with C as a center 
and iCas a radius, then joining their points of intersection by a 
straight line tp; the intersection of this line with CW being the 
point S'. This is the method given by Klein. 

Demonstration. — Since t is on the first of the two circles, 
CS'XCW = (Ct) 2 , but Ct= AC. :. CS'XCW = (AC)\ Q.E.D. 
The third method is shown in Fig. 78 and consists in finding by 
the ordinary construction of plane geometry a point L on the rod 




Fig. 78. 

such that CL X CW = (OC) 2 . Then from L draw a perpendicular 
to CW, meeting OW in M, and from M draw a perpendicular to OW } 
which will meet CW in S'. This is the method given by Dennett. 



104 



DYNAMICS OF MACHINERY 



Demonstration. — Since CL X CW = CO 2 , and since 
CL = CS' + LS' = CS' + MS'(~)== CS' + WS' (S 

we obtain by substitution 

CS' X CTF (f^)V (C#) 2 = (CO) 2 . 

.\ CS' XCW^ = CO 2 - CR 2 = OR 2 . 

i /cw\ ) 2 
/. C£'xCTF = JOflQ^Jj =(AC) 2 . Q.E.D. 



Throws of the Rod in Magnitude and Direction. 
Let W = weight of the rod, and a = angular velocity of the crank 



in radians per second, so that a = 



2tN 
60 



where N = number of 




revolutions per minute. Then if, in Fig. 79, 
OC be laid off to scale to represent a 2 r, and 

CW, (a 2 r) - = a 2 l, then we shall have 



Throw in a direction parallel ) _F,.« 
to path of crosshead pin )~~~g^' m 

Throw perpendicular to path ( _ W ( . n 

w 

= — (01). 
9 



of crosshead pin 
Resultant throw 



GRAPHICAL CONSTRUCTION 105 

If, on the other hand, OC be laid off to scale to represent 

W (W \l w 

— a 2 r, and CW to represent [ — a 2 r - = — a 2 l, then we shall have 
9 \9 Jr g 

Throw in a direction parallel to path of ! _ j 

crosshead pin j 

Throw perpendicular to path of crosshead pin = ij. 

Resultant throw = 01. 

Demonstration. — The resultant throw is always equal to the 
mass multiplied by the acceleration of the center of gravity, as 
has been already shown. 

Point of Application of the Throw Parallel to the Path of the Cross- 
head Pin. 

The distance from W to the point of application of the throw 
parallel to the path of the crosshead pin has been already shown 

t0 be , _ (ji-ft) 

_ _ Mi- P )+h P _ }1 I p 

Xl /, (l- *.) + Mo Xo u p--% u 

Hence the ratio of x 1 to x is the same as the ratio of the accelera- 
tion of the center of percussion with W as center of oscillation, to 
the acceleration of the center of gravity; hence we have 

WT = dd 1 

WG ~ il ' 

Hence to locate T, draw from i a line it in any convenient direc- 
tion, and lay off on it ig = WG, join the point g with I by a straight 
line gl, and from j, the intersection of dd 2 with iG, draw the line jt 
parallel to Ig, cutting it in t. Then will it be the distance from W 
to the point of application of the throw parallel to the path of the 
crosshead pin. 

The point of application of the throw at right angles to the path 
of the crosshead pin is D, the center of percussion of the rod, with 
W as center of oscillation. Hence to find the point of application 
of the resultant throw, draw through T a line parallel to OW, 
and through D a line perpendicular to OW. Then through their 
point of intersection K draw a line KV parallel to 01. The point 
V, where this line cuts the rod, is the point of application of the 
resultant throw. 

Equivalent System of Concentrated Weights. 

We may assume any point on the rod, as W, for center of oscil- 
lation and determine the corresponding center of percussion D. 
Then if we substitute for the weight of the rod two separate weights, 



106 DYNAMICS OF MACHINERY 

one concentrated at the center of oscillation, and the other at the 
center of percussion, these two weights being so proportioned that 
their sum is equal to the weight of the rod, and that their com- 
bined center of gravity coincides with that of the rod, we shall have 

1° The sum of the two weights is equal to the weight of the rod. 

2° The statical moment of the two weights about any axis per- 
pendicular to the plane of the paper is equal to the statical 
moment of the rod about the same axis. 

The moment of inertia about an axis through the center of 
oscillation and perpendicular to the plane of the paper is equal to 
the moment of inertia of the rod about the same axis; and hence 
the moment of inertia of the two weights about any other parallel 
axis is equal to the moment of inertia of the rod about the same 
axis. 1° and 2° are evident from the construction. 

As to the moment of inertia of the rod about the axis of oscilla- 
tion W, we have the following: 

Let W = weight of rod, WD = P ; WG = x . 

WXn 

Then the two parts into which we divide the weight will be 

at the center of percussion, and at the center of oscilla- 

P 
tion. Hence, after the concentration, the moment of inertia will be 

(E^y = Wxop = Jf 

or, in words, the moment of inertia is the same as that of the rod. 
Instead of assuming W as center of oscillation, we may, of 
course, assume any other point in the rod, determining the corre- 
sponding center of percussion; but there is no special object in 
doing so, though some constructions are based upon assuming C 
as center of oscillation, and determining the corresponding center of 
percussion. 

Another Construction for Determining the Point V, the Point of 
Application of the Resultant Throw. 

At least half a dozen other constructions have been devised to 
determine the point V, but as, in the opinion of the author, the 
more important points are D and T, it seems to him that the 
chief use of any of these other constructions for determining V 
is that of a check upon the accuracy of the drawing. Hence 
only one of these methods will be given, as follows : 

Substitute for the weight of the rod two properly proportioned 
weights, one concentrated at W, and one at D, then the accelera- 
tion of W is along OW, and that of D is parallel to Od. Hence the 
resultant throw must pass through the intersection I of OW with 
a line Dl parallel to Od. Hence, through I draw a line parallel to 
01, and its intersection with CW will be V. 



GRAPHICAL CONSTRUCTION 107 

Method of Procedure 

In order to determine the effect of the reciprocating parts of a 
steam engine, the following constants must be known: 

1° The combined weight of Wi of the parts which have only a 
reciprocating motion (piston, piston rod, and crosshead). 

2° The distance x of the center of gravity of the connecting 
rod from the crosshead pin. 

3° The distance p of the center of percussion of the connecting 
rod (corresponding to the center of the crosshead pin as 
center of oscillation) from the crosshead pin. 

4° The weight W of the connecting rod 

We then should determine the two parts of the weight of the 
connecting rod, which we may consider as concentrated at the 
crosshead pin and at the center of percussion respectively. 

Denote by Wi = W the portion of the weight of the 

P 
connecting rod which we may consider as concentrated at the cross- 
head pin, and let ~FP 3 = Wi + W 2 - 

or n 
Denote by Wq = W — the portion of the weight of the connect- 

p 
ing rod which we may consider as concentrated at the center of 
percussion. 

Also let A — area of piston in square inches. 
r = length of crank in feet. 
I = length of connecting rod in feet. 
N — number of revolutions per minute. 

2ttN irN . , , , 

a = Rn = -jtjt = angular velocity ol crank. 

g = acceleration due to gravity = 32.16 feet per 
second. 

Compute the quantity 

f = Ws 2 = 4tt 2 TF sN 2 r _ 1.2271 W 3 N 2 r 
J ~gA ar 3600 gA 3600 A 

which is the number of pounds per square inch of piston area 
corresponding to the force required to produce in a unit of time 
(a second) the (radial) acceleration of the crank-pin center, in a 
weight Wz placed at the crank-pin center. 
Then we compute 

/ = 1.2271 W 3 N 2 r 

S 3600 SA ' 

where S is the scale of the drawing (pounds per square inch of 
piston area to the linear inch) , and OC is measured in inches. Using 

this value for OC, and for CW the value of - (OC), and with the 



108 



DYNAMICS OF MACHINERY 



crank angle COW for which the results are desired, we construct 
a diagram like that shown in Fig. 80. 



Laying off 



WG = x ~- and WD = p (—-) 




Fig. 80. 



Then we shall have 

OC = length of line representing the pressure per square inch 

of piston area corresponding to the accelerating force 

of weight Wz at the crank pin. 
OS = the same, corresponding to the accelerating force of weight 

W3 at the crosshead pin. 
Od 2 = the same, corresponding to the component, parallel to 

the path of the crosshead pin, of the accelerating force 

of weight W3 at D. 
Odi = the same, corresponding to the component, at right angles 

to the path of the crosshead pin, of the accelerating 

force of weight Wz at the center of percussion. 

But if we represent by Wo, as stated above, the portion of the 
weight of the connecting rod which we may consider concentrated 

f Wc 
\W V 

representing the pressure per square inch of piston area corre- 
sponding to the component, parallel to the path of the crosshead 
pin, of the accelerating force of W at the center of percussion. 

— ) Odi = the same, corresponding to the component, at right 

angles to the path of the crosshead pin, of the accelerating force 

of weight Wo at the center of percussion. 

F 
Hence the value of -7 (see page 48), the equivalent force per 

square inch of piston area, which if applied at the piston would 
produce the actual rotative effect due to the throw parallel to the 



/Wq\ 

at the center of percussion, we shall have ( ^- J Od 2 = length of line 



GRAPHICAL CONSTRUCTION 



109 



fW 



path of the piston, is OS + ( = J Od 2 , and the value of F 3 will 



be^JOd, 



Rotative Effect. 



The rotative effect of OS is 



that of 



and that of 






[ 



IS 



IS 



W* 



Od 



VQL\ 

2 \\ocr 

~\/QM\ 

1 \oc) 



These rotative effects can be deter- 
mined by computation, or by graphical 
processes, as follows: 



1 ( 



Rotative effect of OS : 

Make VS and VA (Fig. 81) m 
equal respectively to OS and OA 
of Fig. 80; then lay off VC equal 
to OC of Fig. 80, and draw CS, 
and parallel to it AT, then will 
VT be the rotative effect desired. 

Wo 

Make OC, OK, and OL (Fig. 
81) respectively equal to OC, 



2° Rotative effect of 



Od, 




Fig. 81. 



f^\ Od 2 , and QL of Fig. 80. Draw CK, and parallel 

to it LX; then (OC) (OX) = (OK) (OL), and hence OX is 
the rotative effect desired. 

Wo 



Rotative effect of (^ jOdii 

Make OC, 01, and OM equal respectively to OC, 
{^\Od h and QM of Fig. 80. Draw MY parallel to C7. 

Then OF is the required effect. 

As the rotative effects parallel and perpendicular to the path 
of the crosshead pin act in opposite directions, therefore 

XY = OX - OY 

is the resultant rotative effect of the accelerating force of 
the weight Wo at the center of percussion. 



110 



DYNAMICS OF MACHINERY 



5° The total rotative effect due to the acceleration of the recip- 
rocating parts is VT + XY. 

6° On the other hand, if the rotative effect due to the accelera- 
tion along the line of dead points is all that is desired, it is 

VT + OX. 

7° To find the rotative effect due to the acceleration of the 
reciprocating parts when the crank angle is 90°, draw CS 
(Fig. 82) perpendicular to CW. Draw Ddi parallel to 

OW and erect the perpendicular 
dkdi. Then as the rotative effect 
due to the throw at right angles 
to the path of the crosshead pin 
is zero, and the rotative effect of 
the accelerating forces parallel to 
the path of the crosshead pin 
have the same values as the forces 




>w 



(Wo 
\Ws 



OS and I t==t )Od 2 respectively, it 



Fig. 82. 



is not necessary to construct tri- 
angles as in Fig. 81, but, in con- 
structing our diagram of rotative effects, we can lay off 

W n 
OS+ V ^Od 2 , 
Wz 

as the rotative effect due to the acceleration of the recipro- 
cating parts when the crank angle is 90°. 

After finding the rotative effect in the above manner for 
crank angles 30°, 60°, 90°, 120°, and 150°, we can construct a dia- 




Fig. 83. 

gram of rotative effects (Fig. 83) from which the rotative effect 
due to the acceleration of the recipro- 
cating parts can be obtained for every 
ten degrees. 

It will, also, be found convenient to 
construct a diagram like Fig. 83, using 
values of OA (Fig. 80) as ordinates for 
crank angles 30°, 60°, 90°, 120°, and , _ 
150°. This gives values of OA for every 
ten degrees, and by making OP in 
Fig. 84 equal to the actual steam pres- 
sure (taken from the true stroke card) OCi = OC of Fig. 80, 
and OA equal to the ordinate from this diagram, we find OZ to 




Fig. 84. 



BALANCING 



111 



be the rotative effect due to the steam pressure. This should 
be found for every ten degrees, and corrected by adding or sub- 
tracting the values for the corresponding crank angles from Fig. 83. 



IF 



Balancing. 

Balancing the Action of the Reciprocating Parts in a High-speed 
Engine by Means of Counterweights. 

In the case of very slow-speed engines, both the horizontal and the 
vertical throw of the reciprocating parts are small, and hence it is 
not of any great consequence that they should be balanced; but, 
as the speed increases, this matter becomes of greater and greater 
importance, in order to avoid injurious strains in the frame, as 
well as on the foundation, of stationary engines; on the hull of the 
boat in marine engines; and, in the case of locomotives, to avoid 
injury to the track and the roadbed in consequence of the vertical 
throw, and to avoid undue strains in the frame and in the draw- 
bars, and uneven riding, whether in the cars or on the locomotive, 
on account of the horizontal throw. 

Balancing Revolving Masses. 

Suppose (Fig. 1) we have a horizontal shaft revolving in its bear- 
ings A and B carrying at C a truly turned and uniformly thick 
homogeneous circular disk, then 
the only pressures on the boxes 
are those due to the weight of 
disk and shaft and are constant. 

Suppose now we add at D, 
where CD = r, a, weight W in 
the plane of the disk (this can 
be done by bolting \ W on each 
side at D); then, if a = angular 
velocity in radians per second, a 

W 

centrifugal force F = — a 2 r will 

9 

be developed, which acts along the 

line CDF outwards. This force, 

when combined with the weight of the shaft and disk, and the 

weight W, causes pressures on the bearings varying in amount 

and in direction throughout each revolution, thus bringing loads 

on the bearings variable in amount and in direction. This throw 

can be balanced by placing at E, directly opposite D, and where 

r 
CE = r h a weight TFi, such that Wi = W -, since the centrifugal 

W voPt\ "WcPt 

force due to this latter weight is = = F and acts 

9 9 

directly opposite to the centrifugal force due to W at D. 



A 



i 



Fig. 85. 



112 



DYNAMICS OF MACHINERY 



A rotating body, as a rotating part of a machine, is said to be in 
standing balance when its center of gravity is in the axis of its 
shaft. Whether this is the case, may be ascertained by resting the 
shaft carrying the revolving body in various positions on a pair of 
parallel horizontal rails. If the center of gravity is in the axis, 
the body will rest in any position in which it is placed, but if it 
is not, the body with the shaft will roll over until the center of 
gravity is vertically below the axis of the shaft. 

In the latter case, standing balance may be secured by adding 
weights at suitable points above the axis symmetrically placed 
with reference to the vertical plane containing the axis, determin- 
ing the magnitude of these weights by trial, since we do not know 
the actual position of the center of gravity of the unbalanced 
body and its shaft combined. A rotating body, however, which 
is in standing balance, may not be in running balance, as there 
may exist a centrifugal couple, tending to turn the axis around a 
line perpendicular to it, as will be explained in the two following 
cases. (See Figs. 86 and 87.) In Fig. 86 the body consists of 




Fig. 86. 



Fig. 87. 



two equal weights, each equal to W at E and F respectively, 
attached to the shaft AB, by rigid and weightless wires CF and 
DE, the weight of the shaft being also disregarded. 

Let CF = DE = r, and let CO = DO, and AO = BO; then, 
when this combination revolves with an angular velocity a radians 

W 

per second, the centrifugal force of each weight is — a 2 r, and as 

these form a couple whose arm is CD, the moment M of this cen- 

(W \ 
trifugal couple is ( — a 2 r)CO, and hence, notwithstanding the fact 

that the combination is in standing balance, its center of gravity 
being at 0, when running it is subjected to a centrifugal couple M 
which is balanced by a couple formed by the pressures on the 

M 

bearings. The reactions of the bearings are each -7-5 , that at A 

being parallel to and in the same direction as the centrifugal force 
at E, while that at B is parallel to this reaction, and opposite in 
direction. 



BALANCING 113 

Observe, also, that the principal axes of inertia are respectively 

FE, GH (when GOF = -J and an axis through perpendicular 

to the plane containing FE and GH. Had the axis of the shaft 
coincided with any one of these three, there would have been no 
centrifugal couple. 

Fig. 87 shows a closed cylindrical and symmetrical drum, 
mounted on the shaft AB along its axis. With this alone, the 
body is in standing balance, and also in running balance. 

In this case there would be no centrifugal couple, and the 
pressure on the bearings would be only those due to the weight of 
the body. Now suppose that equal weights W are added at F 
and E respectively, where CF = DE = r; then the body is in stand- 
ing, but not in running, balance, and the centrifugal couple will 

be M =( — a 2 r)CD, and this will cause pressures in opposite 

M 

directions on the bearings, each equal to -r-- • 

, Observe that the principal axes at can be found by the methods 
already given under moments of inertia. They are (a) an axis 
perpendicular to the central plane FCDE ; (b) an axis passing 
through 0, inclined to OC at an angle less than the angle COF; 
and (c) an axis at right angles to (a) and (6). 

Had the axis of rotation coincided with one of these, there would 
have been no centrifugal couple. 

In order to balance the centrifugal couple, weights must be 
placed at such points as to produce a centrifugal couple of equal 
magnitude, and opposite sense. This can be done by placing 
weights, each equal to Wi, at G and H respectively, where CG = 
DH = ri, provided Wi is so chosen that Wifi = Wr, i.e., provided 

CASE II. 

Suppose we have a horizontal shaft revolving in its bearings 
at A and B, carrying a disk at C, and another at D, both of which 
are uniform in every way, so that the combination of the shaft 
and disks is not only in standing but also in running balance, Fig. 
88. Suppose that to the left of the disk C is attached a weight W, 
whose center of gravity is at E in the vertical plane EFG. 

Let EF = r, FC = e, CD = s. 

Then if this system revolves with an angular velocity a, there will 

be developed an unbalanced centrifugal force along the line FE 

Wa 2 r 
equal to , and hence the system is neither in standing nor in 

running balance. 



114 



DYNAMICS OF MACHINERY 



Of course, it could be balanced by placing a weight Wi at G, 

Wr 
where FG = n and where Wi = — , but as in many practical 

cases this is not feasible, another method will be explained, viz., 
balancing the unbalanced centrifugal force by means of two counter- 
weights, one in the disk C, and another in the disk D, that in the 



clj 



i — i 
»■•-■! H 



t^F 



B L 



IP- 



l 



Fig. 88. 



disk C being so placed as to cause a centrifugal force acting in 
a direction parallel and opposite to the centrifugal force of W, 
while that in the disk D must be so placed as to cause a centrif- 
ugal force, parallel and in the same direction as that of W. 
Thus, if the counterweight in disk C be W c placed at H where 
CH = r, and if the counterweight in disk D be Wa placed at K 
where DK = r, we shall have, taking moments about D, 



W c s = W(s + e); 


... Wc = W s + e t 

s 


and, taking moments about C, 




W d s = W e ; 


s 


Moreover, W c - 


■W d = W. 



On the other hand, if the counterweight in disk C be placed at Hi 
where CHi = r 2 , and if its magnitude be denoted by W c ', we shall 
have 



W c ' = W e -, 
r 2 



BALANCING 



115 



and, if the counterweight in disk D be placed at K x where DK X = 
r 3 , we shall have 

wi = w d -- 

case ni. 

Suppose we have the same shaft and disks as in case II, but 
that the unbalanced weight W is at E between the disks. 



rti 



- ES- 



C 



H 



1 IF 



m^ 



e^ 



St ' g 

i 



Fig. 89. 



K 






4J 



T ** 

3 



Let CD = «, CF = e, F# = r, CH = DK = r, CHi 
MCi = r 3 . 

W c = counterweight in disk C if placed at #. 
Wd = counterweight in disk D if placed at K. 
Wc = counterweight in disk C if placed at Hi. 
Wd = counterweight in disk D if placed at K\. 

Then we have, taking moments about D, 



ri> 



W c s = W(s-e); 


.-. w c -w* * 

5 


and taking moments about C, 




W d s = We; 


.-. w d = w-- 

s 


Moreover, W c + W d 


= w. 


We also have 





W c ' = Wc L and WJ = W d - 
T2 r 3 



116 DYNAMICS OF MACHINEEY 

The principles explained above find their application in the 
balancing of the action of the reciprocating parts of steam engines 
whether stationary, marine, or locomotive. 

Balancing the Action of the Reciprocating Parts in a Locomotive. 
The main objects to be accomplished by counterweights are 

1° To avoid injury to the track and roadbed. 

2° To avoid the decrease of weight on drivers, and hence of 

the tractive force when the vertical throw is upwards. 
3° To provide comfort in riding. 
4° To avoid injury to the locomotive itself. 

Of these, the fourth can be left out of consideration, for it will 
be taken care of when the other three are provided for. 

The first and second depend upon the degree of perfection with 
which the vertical throw is balanced, while the third depends upon 
the perfection with which the horizontal throw is balanced. The 
piston, piston rod, and crosshead have no influence upon the 
vertical throw, whereas they have a great deal of influence on 
the horizontal throw. 

Vertical Throw. 

Consider the vertical throw due to the connecting rod only of 
one cylinder of a horizontal engine. We have for the vertical throw 
when the crank is at right angles to the line of dead points, 

p Sb „ 
T — a 2 r, 

where 

Sb = weight of crank end of connecting rod. 
I = length of connecting rod, center to center, 
p = distance from crosshead pin, center to center of percus- 
sion. 
r — length of crank. 
a = angular velocity of crank in radians per second. 

This vertical throw is the same as the centrifugal force of a 
weight y Sb placed at the crank-pin center. 

The weight =■ Sb is often called, on this account, the revolving 

weight, and many have assumed for it some definite fraction of 
the weight of the rod, some assuming one-half, some two-thirds, 
and others some different proportion. 

CASE I. 

Consider a two-cylinder simple locomotive with only one pair 
of drivers, the axes of the cylinders being outside the wheels. In 
order to balance the vertical throw of the reciprocating parts 



BALANCING 



117 



belonging to one cylinder, we must balance, by means of counter- 
weights in the two driving wheels, the two following weights: 

(a) w c = weight of crank. 

(6) P r = jSb + w p = revolving weight, where w p = weight 
of crank pin. 

We can either compute the counterweights suitable to balance 
each of these two weights separately, using the method already 
explained, and then combine these counterweights, or else begin 
by finding the center of gravity of the resultant unbalanced 
weight whose magnitude is P r + w c , and then compute the suita- 
ble counterweights for this resultant unbalanced weight, using the 
methods already explained. 

The following formulae are deduced on the assumption that each 
crank has been already balanced in its own wheel, and hence that 
the only weights to be considered are the two weights P r , one on 
each side. 




Fig. 90. 

To balance the upper revolving weight P r , let 

CD = s, CF = e, FE =r, CH = r, DK = r. 
Then if we let 

W c = counterweight required in wheel C at H, 
Wd = counterweight required in wheel D at K, 

we shall have 



118 DYNAMICS OF MACHINERY 

and W d = P r -' 

s 

Hence in wheel C we shall have a counterweight W c opposite its 
crank, and, since the cranks are at right angles to each other, 
we shall have in wheel D a counterweight Wd at right angles to 
its crank. 

On the other hand, in order to balance the lower P r we shall 
have to place in wheel D opposite its crank a counterweight equal 

o I p 

to P r , and in wheel C at right angles to its crank a counter- 

weight equal to P r - • 

o 

Hence in each wheel we shall have two counterweights, one 

S —1— p 

equal to W c = Pr opposite the crank, and one equal to 

o 

Wd = P r - at right angles to the crank. 

These four counterweights, two in each wheel, will accomplish 
the balancing, and will be convenient to use. 

If, however, for any reason it is preferred to use only one coun- 
terweight in each wheel, its magnitude will be 



W c ' = VWc 2 + Wd 2 , 

and the angle ^ which will be formed by the line drawn from its. 
center of gravity to the axis of the shaft with the line from the: 
center of the shaft to the center of gravity of W c , will be given 
by the equation 

* , W d 
tan + = W ~ 



CASE II. 

If we take the case of an ordinary eight-wheel engine, 4 — 4 — 0,. 
and still assume that the cranks are all independently balanced 
in their respective wheels, we have, besides the counterweights 
in the main driver, deduced in case I, to balance also in the main 
driver one -half the parallel rod, and in the rear driver we have 
to balance one -half the parallel rod plus the rear crank pin (see 
Fig. 91). 

Let 

CD - C 2 D 2 =s, CF = e, CF X =e x = C 2 F 2 , FE = Pi#l 
= F 2 E 2 =r. 

p Sb 

I 9 
R = weight of parallel rod. 



P r = 7 — a 2 r + w p , as before. 



BALANCING 



119 



^ R + C = weight of one-half parallel rod plus weight of rear 
crank pin. 
W c = counterweight in forward driver opposite crank pin. 
Wd = counterweight in forward driver at right angles - to 

crank pin. 
W c ' = counterweight in rear driver opposite crank pin. 
W/ = counterweight in rear driver at right angles to 
crank pin. 




Fig. 91. 



Then we have when 



CH =DK = C 2 H 2 = D 2 K 2 = r, 

s 2 s 

W d =P r - + ] ) R e -±> 

s 2 s 



MV = [|fi + C 



/ 8 



The two last are slightly inaccurate because the center of 
gravity of the rear crank pin is not at E 2 , but as a rule it will be 
unnecessary to perform the additional work needed to introduce 
this refinement. When the points HiKiH 2 and K 2 are so chosen 
that we do not have CH = DK = C 2 H 2 = D 2 K 2 = r, the above 
values should be multiplied by r and the product divided by the 



120 DYNAMICS OF MACHINERY 

corresponding value of these quantities. For the case of inside 
cranks the same principles apply. If we desired to balance the 
whole horizontal instead of the whole vertical throw, we should 
adopt a similar method of procedure. It will not do, in practice, 
to balance only the vertical throw, unless the locomotive happens 
to be of such a kind that, as in the case of the balanced compound, 
the whole horizontal throw would be thereby balanced, at the 
same time. 

In any other class of locomotives, balancing only the vertical 
throw would have for result so great a plunging that, not only 
would the riding be very uncomfortable, but also that injurious 
strains would be developed in the engine and in the drawbars. 

Hence it is necessary (a) to balance the whole vertical throw 
at each wheel, in that wheel, and (6) to add an additional balance 
known as the excess balance. 

This excess balance is for the purpose of balancing partially 
the horizontal throw. Moreover, this excess balance is usually 
divided equally between the driving wheels. 

The rules followed by builders of locomotives generally reduce 
more or less approximately to the following, viz. : 

Counterweight the vertical throw of each wheel by a counter- 
weight in the wheel, and then divide the excess balance equally 
between the drivers. As to how large this excess balance should 
be, there is more or less difference of opinion. 

One rule proposed, but not always followed, is to divide the 
excess balance that would be required if the whole horizontal throw 
were to be balanced by the number of pairs of drivers plus one, 
and to balance that amount in each driver, thus leaving that 
amount unbalanced. 

In a paper presented to the American Society of Mechanical 
Engineers (see Vol. XVI of the Transactions) by Mr. David L. 
Barnes, he says: "So far as the locomotive itself is concerned, the 
balancing is practically perfect when the balances are placed in 
the wheels opposite the crank pins, and when all the revolving 
parts are balanced, and not more than 100 pounds of reciprocating 
parts for light engines, and 300 pounds for heavy engines, are left 
unbalanced." He also says: "The heavier the locomotive, the 
greater is the amount in pounds of the reciprocating parts that 
can remain unbalanced without causing the locomotive to shake 
in nosing and plunging more than can be permitted. It is not 
the percentage of the total weight of the reciprocating parts that 
should be considered in selecting the excess balance, it is the actual 
weight in pounds that can remain unbalanced without shaking 
the engine too much. If one-third of the weight of the reciprocat- 
ing parts weighing 600 pounds can remain unbalanced, then, if 
these parts can be reduced to weigh but 400 pounds, one-half 
can remain unbalanced, and the excess balance will be needed 
for but 200 pounds, instead of 400 pounds, of reciprocating 
weight." 



BALANCING 121 

Counterweighting a Balanced Compound Locomotive with Two Inside 
and Two Outside Cranks, All on the Same Axle. 

This case will be given as an additional example. 

The balanced compounds have four cylinders, two high- and two 
low-pressure, the cranks for the high-pressure cylinders being often 
inside, and those for the low-pressure cylinders outside. In such 
cases the cranks of one high-pressure and of the corresponding 
low-pressure cylinder usually make an angle of 180° with each 
other, while the other two cranks make angles of 90° with the first 
pair. 

In some such locomotives the two high-pressure cranks are 
on the rear driving axle, while the two low-pressure cranks are on 
the forward driving axle; while in others the four cranks are all 
on one driving axle. 

The latter case will be considered here. 

It is usually necessary to consider only the revolving parts, as 
the remainder of the reciprocating parts are so constructed as to 
nearly balance each other. 

Hence in the following example we will consider only the revolv- 
ing parts. 

Example. — Given a balanced compound locomotive, the axes 
of the two high-pressure cylinders being inside, and those of the 
low-pressure outside, the driving wheels. Let the dimensions be 
as shown in the figure, let the weight of the revolving parts be 500 
pounds for each low- and 1200 for each high-pressure cylinder. 
The arrangement of the works is shown in the figures. To balance 
the revolving parts at the two left-hand crank pins as shown in the 
figures (see Figs. 92 and 93). 

The calculations are as follows: 

We = 500 £■# = 625 pounds = counterweight in left wheel, in 

front of left outside crank pin. 

Wd = (500) $■# = 125 pounds = counterweight in right-hand 

wheel at right angles to line 
of right-hand cranks. 

W c " = (1200) %$ = 900 pounds = counterweight in left wheel op- 
posite the outside crank pin. 

Wd" = (1200) J£ = 300 pounds = counterweight in right wheel at 

right angles to line of right- 
hand cranks. 

Hence we have, in left-hand wheel in line with the outside left 
crank pin: 

W e = 900 — 625 = 275 pounds = counterweight in left-hand 

wheel in line with outside 
left crank. 

Wd = 125 + 300 = 425 pounds = the counterweight in the left 

wheel at right angles to 
line of cranks. 



122 



DYNAMICS OF MACHINERY 




A C 



~DK 



s- 



V 



OS 

IS 



8:2 




«qiszj s?M 



STRESSES IN CONNECTING RODS 



123 



Moreover, we have to balance also the revolving parts at the two 
cranks on the right-hand side of the engine. Hence we should 
have in the left-hand wheel two weights, viz., one of 275 pounds 
in front of the left-hand outside crank pin, and one of 425 pounds 
at right angles to line of outside cranks, and similarly two weights 
of 275 and 425 pounds in the right wheel. 



L.P. 



I 



=s- 



H.P. 



=& 



H.P. 



L 



J 



=S- 



L.P. 



^S- 



Right Outside </ 



Left-Inside 




:rj. Left Outside 



Right Inside 



Fig. 93. 

The figures show the location of both counterweights as needed. 
The two weights in each wheel can be replaced by a single 

resultant weight whose magnitude is R = V(275) 2 + (425) 2 = 506 
pounds, located as shown by dotted lines, where tan 6 = f £f .* 



CONNECTING RODS. 

Calculation of Stresses in the Body of the Main Rods of Locomotives. 

Almost all main rods of locomotives belong in one of the two 
following classes, viz. : 

(A) Those in which the width and the thickness of the flanges, 

as well as the thickness of the web, are constant, while 
the depth of the web has a uniform taper. 

(B) Those in which the depth and the thickness of the web, as 

well as the width of the flanges, are constant, while the 
thickness of the flanges has a uniform taper. 

The formulae for each of these two classes will be deduced, an 
I section being assumed. 



* In 1907 Mr. L. H. Frye published an article on this subject in "Recent 
Locomotives." 



124 



DYNAMICS OF MACHINERY 



Formulae for rods of rectangular section and uniform thickness 
can be obtained from those for class (A), by substituting zero for 
the width and for the depth of the flanges. 

Formulae for rods in which the thickness of the web, or the 
width of the flanges, or both, have a taper will not be given, but 
they can be obtained by a method of procedure similar to that 
employed here. 

In deducing the stresses for main rods the stub ends, including 
the straps and brasses will be neglected, since each stub end pro- 
duces in the body of the rod a positive and a negative bending 
moment nearly equal in magnitude. 

In classes (A) and (B), for the body of the rod, will be sub- 
stituted a rod body of uniform taper throughout, having a length 
equal to the distance from crank-pin center to crosshead-pin 
center, the plan of this substituted rod being a rectangle, and its 
elevation a trapezoid, as shown in Figs. 94, 95, 96, and 97. 






^ 



I 



^4 



J 



* | 

7T~F +~ 




Figs. 94 and 95. 



Fig. 94 shows the plan and elevation of a rod of class (A), and 
Fig. 95 those of the substituted rod. Fig. 96 shows the plan and 
elevation of a rod of class (B), and Fig. 97 those of the substituted 
rod. 

In all these figures : 

Let X be the crosshead-pin center. 
Y be the crank-pin center. 
a = width of flange in inches. 
t = thickness of web in inches. 
L = length XY in inches. 
L\ = length XiYi in inches. 
Z/2 = length XX i in inches. 
Lz = length YYi in inches, ,\ L = L\ + L 2 + Li. 



STRESSES IN CONNECTING RODS 



125 



In Figs. 94 and 95: 

Let d = thickness of flange in inches. 

bi = depth of web of rod at Xi in inches. 
Bi = depth of web of rod at Fi in inches. 
b = depth of web of substituted rod at X in inches. 
B = depth of web of substituted rod at F in inches. 

Then we easily derive from the figures 

6 = 6i-(^f^)i 2 and B = B 1 + (^=-^) L-, 



— = Fbr^r 



f 



E 



ii 



3^: 



8|= 
T 






«8 
J*. 



S 






X j Xl 

<-L 2 -*k- 



Q^l 



5 15^ 



> 



Figs. 96 and 97. 



On the other hand, in Figs. 96 and 97: 

Let di = thickness of flange at Xi in inches. 
Di = thickness of flange at Fi in inches. 
d = thickness of flange at X in inches. 
D = thickness of flange at F in inches. 
b = depth of web of rod and of substituted rod in inches. 

Then we easily derive from the figures 

d = di -(^zr 1 ) L * and D = D >+( Dl L 1 dl ) u 

We now proceed to work with the substituted instead of the 
actual rods. Moreover, the dimensions common to both substi- 



126 



DYNAMICS OF MACHINERY 



tuted rods, i.e., that of class (A) and that of class (B), are as 
follows : 

L — length of rod in inches. 

a= width of each flange in inches. 

t = thickness of web in inches. 

Moreover, 

Let r = length of crank in inches. 

w = weight of one cubic inch of steel = 0.2833 pounds 

approximate. 
N = number of revolutions of the crank per minute. 
g = acceleration due to gravity = 386 inches per second. 
a = angular velocity of crank in radians per second. 

= 7riV 3.1416 N 

" a ~ 30 ~ 30 

/ r* 
Let = cos -1 v/ 1 — y^ = angle made by rod with line of dead 

points, when crank angle = 90°. 

Observe that the vertical throw of the rod induces a vertical dis- 
tributed load upon it, which is greatest at Y and which decreases 
gradually to zero at X. 

The figure below exhibits this distribution. 

We will next proceed to find 
the expression for the bending 
moment at any section whose dis- 
tance from X is e, due to the 
vertical throw when the crank 
angle is 90°. Observe that the 
bending moment due to the hori- 
zontal throw is neglected because 
it is small. 
Inasmuch as the formulae for the two classes of rods will differ, 
we will first deduce those for class (A), and subsequently those 
for class (B). 

CLASS (A). 

Additional dimensions are: 

d = thickness of each flange in inches. 
b = depth of web at X in inches. 
B = depth of web at Y in inches. 
We then have 

1° Area of cross section of rod at distance x inches from X 




Fig. 98. 



2ad + t(b+?-j-^x). 



2° Let F = vertical reaction at X. 

3° Then will F cos equal component of this reaction at right 
angles to the rod. 



STRESSES IN CONNECTING RODS 127 

4° Vertical throw of elementary disk of length dx at distance 
x from X 

w <* 2rx ^2ad + t(b+*^x)ldx. 



g L(- ! "V L 

5° Component of 4° at right angles to the rod 

wa 2 r ^ i > Ji , B — b \) 

lad -{- 1\ o-\ = — x >x cos 6 dx. 



6° Bending moment at section e inches from X 
M =(Fcosd)e-^cosd fj2ad+t(b +^-^x)lx(e-x) dx,, 

/. M = Pe cos - ^cos B \ (2 ad + tb) ~ + t^^ e~ 
/. M = cos \Fe j- \ (2 ad + #>) ■=■ + Z — f— t^ 

L [Q^ ( 6 Li 11 

Inasmuch as, in practical cases, cos is very nearly 1, it will be best 
to put cos = 1, and thus obtain 

M = Fe - Ge 3 - He*, 
where 

= ^12(2^)1 and tf=fj(^> 

To determine F observe that when e = L the bending moment 
becomes zero, and hence FL — GL 3 — HL* = 0; .". F = GL 2 + 
HL 3 . 

Hence, to obtain the bending moment at a distance e from X, 
the crosshead-pin center, we proceed as follows, viz. : 

(a) From the dimensions and speed compute the values of G 
and H where 

O-gfrlW + W and *=iJj<Vi- 

(6) Compute F from the equation F = GL 2 + HL 3 . 

(c) Write out the bending moment at a distance e from X, i.e., 

M = Fe - Ge 3 - He 4 . 

(d) Let Pi = total effective pressure on the piston, and hence 
total force exerted by the piston rod on the crosshead pin. Re- 
solve Pi into two components, one of which (P) acts along the rod,, 
the other at right angles to the guide. Then 

p = ^. 

COS0 



128 DYNAMICS OF MACHINERY 

(e) Let A = area of section at distance e from X, and let <n = 
intensity of pressure at this section due to the action of the force P, 
then 

P 

0"! = T* 



Moreover, we have 



A =2ad--t\b +?-—-^e 



(/) Let y = half the depth of the rod at the same section; let 
7 = moment of inertia of the section, about a horizontal axis, in 
the plane of the section, and passing through its center of gravity; 
and let cr 2 = outside fiber stress at the section due to the throw. 
Then we have 

My 

Moreover, 




V = o l b H — f— e ( + d > 



and if we let b 2 = b -\ ^ — e = depth of web at the section, we 

have 

/ = ^{a (2 d + b 2 y - ab 2 * + tf> 2 3 ;. 

(gr) Then if o- denotes the total outside fiber stress per square 
inch in the rod at this section, we have 

<T = CTi + (72- 

(h) On a piece of cross-section paper plot a curve having for 
abscissae distances along the rod measured from X, i.e., values of 
e, and for ordinates the corresponding values of a. It will only 
be necessary to calculate and plot four values of a, two corre- 
sponding to values of e less than \ L, and two corresponding to 
values of e greater than | L, and all within six inches of the middle 
of the rod, and then a curve can be drawn through the four points, 
and from this curve can be determined the greatest outside fiber 
stress in the rod. 

In carrying out the above stated calculations we have to use some 
value of N, the number of revolutions of the driver per minute. 
This value should be as large as will ever be attained in practice, 
whether by design or by accident. The author has been accus- 
tomed to use iV = 375. 

As to the value to be used for P x , there may be room for consid- 
erable difference of opinion. The author has generally used the 
product of the area of the piston by one-half the boiler pressure 
per square inch. 



STRESSES IN CONNECTING RODS 129 

CLASS (b). 

Additional dimensions: 

d = thickness of each flange at X in inches. 
D = thickness of each flange at Y in inches. 
b = depth of web throughout in inches. 

We then have 

1° Area of cross section of rod at distance x inches from X 

= 2a(d-\-~^x) + tb = (2ad + tb) +2a^x. 

2° Let F = vertical reaction at X. 

3° Then will F cos = component of this reaction at right 

angles to rod. 
4° Vertical throw of elementary disc of length dx at distance x 

from X 

wa 2 r x { ( . D - d ) 

= y \ (2 ad + tb) + 2 a — ^ — x \ dx. 

9 L ( L ) 

5° Component of 4° at right angles to the rod 

wa 2 r ( ._ 7 . ,,* , _ D — d ) _ , 

(2 ad + tb) + 2 a — f — x > x cos 6 dx. 



gL r ' ~' ' " L 

6° Bending moment at section e inches from X. 

M=(F cos 0) e - ^ -cos I ] (2 ad + tb) 
gL Jo ( 

-\-2a — j — - x [ x (e — x) dx. 
L } 

** n n wa 2 r n l, rt 7 , j7 .v ex 2 . _ D — d ex 2 
:. M = Fe cos j- cos \ (2 ad + tb) -^ + 2 a — ^ 5- 

r 3 D — d T* ) e 

-(2ad + tb)^-2a^ T ^^ I- 

.'. M = cos IVe - ^ j (2 ad + tf>) |' + 2 a ^-^ ^ |1 . 

Inasmuch as in practical cases cos is nearly 1, it will be best 
to put cos = 1, and thus we obtain 

M = Fe - Ge* - He\ 
where 

G = J^L\ 2 (2ad + tb)\ and H = ^{2 «^fA 
12 gL I ) 12tfL\ L j 

To determine JF, observe that when e = L the bending moment 
becomes zero, and hence 

FL - GU - RV = 0; 
... f = GL 2 + #L 3 . 



130 DYNAMICS OF MACHINERY 

Hence to obtain the bending moment at a distance e from X, the 
crosshead-pin center, we proceed as follows, viz. : 

(a) From the dimensions and speed compute the values of G 
and H where 

= J^L\ 2 (2 ad + tb) { and H = ^(2«^V 
12 gL ( ) 12 gL \ L ) 

(b) Compute F from the equation F = GL 2 + HL\ 

(c) Write out the bending moment at a distance e from X, i.e., 

M = Fe - G e * - He 4 . 

(d) Let Pi = total effective pressure on piston, and hence total 
force exerted by the piston rod on the crosshead pin. Resolve 
Pi into two components, one of which (P) acts along the rod, and 
the other at right angles to the guide. Then 

cos 

(e) Let A = area of section at distance e from X, and let ai = 
intensity of stress at this section due to the action of the force P. 

Then P 

,i= z - 

Moreover, we have 

A = (2ad + tb)+2a I ^^e. 

(/) Let y = half depth of the rod, at the same section, and let 
/ = moment of inertia of the section, about a horizontal axis, 
in the plane of the section, and passing through its center of 
gravity. Let o- 2 = outside fiber stress at the section due to the 
throw. Then we have 

My 
I ' 

Moreover, y = -b + Id -\ j — e), 

and if we let 

d 2 = d -f = — e = depth of one flange at the section, 

Lj 

we have 

I = T V \a (2 d 2 + b) s - a6 3 + tb*}. 

(g) Then if a denotes the total outside fiber stress per square 
inch in the rod at this section, we have 

(7 = Ci + (72. 

(h) The remainder of the method of procedure is identical with 
that in the case of class (A) and will not be repeated here. 



*,= j 



STRESSES IN SIDE RODS 131 



SIDE RODS. 



Calculation of Stresses in Side Rods of Locomotives when there 

is No Knuckle Joint. 

Let W = weight of rod minus weight of stub ends in pounds. 

r = length of crank in inches. 

L = length of rod in inches, center to center of crank 
pin. 

g = acceleration due to gravity = 386 inches per second. 

A = area of section in square inches. 

/ = moment of inertia of section of rod, about a hori- 
zontal axis lying in the plane of the section, and 
passing through the center of gravity of the section, 
units being inches. 

y = distance from above stated axis to top or bottom 
of section = one-half entire depth of section in 
inches. 

N = number of revolutions of crank per minute. 

a = angular velocity of cranks in radians per second, 
hence 

_2irN _7rN 

a ~ 60 ~ 30 ' 

F — total throw of rod in pounds. 

d = deflection at center of rod due to centrifugal force in 
inches. 

Mi = bending moment at middle section due to the centrif- 
ugal force only, in inch-pounds. 

P = total force transmitted through the rod. 
M 2 = Pd = bending moment caused by P in consequence 
of deflection d, in inch-pounds. 

M = Mi + Mi — total bending moment at middle section. 

<ti = outside fiber stress in pounds per square inch due to 
Mi. 

<r 2 = outside fiber stress in pounds per square inch due to 
bending moment M 2 . 

p 

a 3 = — = stress in pounds per square inch due to P. 
A. 

a = (7i + (72 + 0-3 = greatest stress in rod in pounds per 

square inch. 
E = modulus of elasticity of material of rod in pounds 

per square inch. 

We then have that 

Wa 2 r 
F = 

g 

The centrifugal force may be considered, with a sufficient degree 



132 DYNAMICS OF MACHINERY 

of approximation, as a uniformly distributed, transverse load on 
the rod. Hence 

, , FL 1 WcfrL 

■M- 1 — • = ~ 



Cl 



d = 



8 8 g 
Miy __ 1 Wa 2 rL y 

384 EI " 3072 ##/ 



We also have 


0"2= j t 
P 




Hence 


0" = 0"i + 0"2 ~T ^3 


Mil/ M 2 y , P 



As to the value to be used for N, it should be the same as that 
used in the case of the main rod. If N = 375 is used in one case, 
it should also be used in the other. 

As to the value to be used for P, there is room for considerable 
difference of opinion. The author would suggest in the case of 
an ordinary eight-wheel locomotive, 

where 

di = diameter of piston in square inches, 



p = 1 Mi* 
2\ 4 



and 



p = one-half the boiler pressure in pounds per square inch. 



CRANK SHAFTS AND OTHER MOVING PARTS. 

In designing, and in determining the stresses in pistons, piston 
rods, connecting rods, crank pins, cranks, and crank shafts, it is 
necessary to take into account the action of the reciprocating 
parts, and especially the rotative effect, in order to determine 
correctly the forces acting upon them. 

To determine the greatest stresses in most of these parts, it 
will often be sufficient to compute them for the crank angle when 
the rotative effect is greatest, although it may be necessary in 
certain cases to determine the stresses that arise when the crank 
is on the dead point. 

In the cases where the engines are of the center-crank type, 
and especially in multiple-cylinder engines, whether they are 
multiple-expansion or not, the calculations generally involve more 



CRANK SHAFTS AND OTHER MOVING PARTS 



133 



complexity, as there are usually more forces to be reckoned with 
than in the case of a single-cylinder side-crank engine. 

The illustration given will, therefore, be of the former class. 
Assume a two-cylinder engine of the center-crank type (Fig. 99), 
the cranks being at 90° to each other, crank bcdef leading, and 
assume the flywheel to be at q, and the power to be taken off at 
the end near q. Assume that the alignment is such that there is 
no bending moment in the portion gh of the shaft, or else that it is 
so small that it may be left out of account. 




Fig. 99. 



In order to discuss the forces acting in the crank klmno and the 
portion of the shaft between h and q, assume the crank angle of 
the crank klmno to be that at which the greatest rotative effect 
occurs. Corresponding to the given crank angle, there is a certain 
driving moment which we will call R, which is transmitted through 
gh, and hence through crank klmno, whose magnitude can be 
determined from the rotative effect of the cylinder to which crank 
bcdef belongs. If r = crank length, and P = force exerted by web 

kl on the crank pin at I, then P = -j , and, moreover, it acts in a 

direction perpendicular to the crank and to the pin. 

This force P = Ir in the figure acts on the crank pin at I, and 
since the crank pin may be considered as fixed at the ends, and 
hence having a point of inflection at the middle, we have a bending 

moment at I, and another at n, each equal to P ~ • 

There remains to consider the force exerted by the connect- 
ing rod on the crank pin at m. This force may be found as 
follows : 



134 DYNAMICS OF MACHINERY 

Lay off ms — rotative effect, and by constructing the rectangle 
mstum we have mt, the force exerted by the rod on the pin. Hence 
we have: 

Force P acting at I in direction perpendicular to crank and 

crank pin. 
Force mt exerted at m, found from ms, the total rotative effect. 

From these the greatest stress at n can be found. Knowing 
these forces, and considering the portion of the shaft from h to 
q, we have in addition the weight of the flywheel at q, also the 
reactions in the boxes. 

The above will serve to show the bearing that the action of the 
reciprocating parts, and especially the rotative effect, has on the 
design of and the determination of the stresses in the moving parts 
mentioned above. 



CHAPTER IV. 
GOVERNORS. 

The function of a governor is to control the speed of a motor 
by varying the amount of energy supplied to it. 

Thus, in the case of some water wheels, the governor operates 
a clutch, a shield, or some other device, so arranged that it throws 
into or out of gear the mechanism (driven by the wheel itself) 
which opens and closes the gate; while, in the case of other water 
wheels, it controls a valve which sets in motion or stops an auxil- 
iary motor by means of which the gate is opened or closed. 

In the case of a windmill, it varies the position of the blades, and 
thus controls the amount of energy imparted to the wheel by the 
wind. 

In the case of a steam engine, it regulates the amount of steam 
supplied to the cylinder at each stroke, and its pressure, either by 
varying the opening of the throttle valve, or else by varying the 
position of the cut-off gear, and therefore the portion of the stroke 
during which steam is admitted to the cylinder. 

In the case of a steam turbine, the governor controls either the 
position of the steam-admission valve, or the number of nozzles, 
and hence the cross section of the steam passages, or the time of 
admission, or else, in cases of overload, it operates valves which 
admit steam at boiler pressure, at various points of the path of 
the steam in the turbine. 

In the case of a gas engine, the methods by which the governor 
controls the supply of energy may be classified as follows, viz.: 

(a) Hit-or-miss regulation. In this case the inlet valve is kept 

closed, and the charge is omitted for one or more firing 
strokes when the speed increases above the normal. 

(b) Regulation by varying the quality, the quantity, or both, 

of the explosive mixtures. In this case, the amount of 
gas, the amount of air, or both, or the amount of the mix- 
ture, is varied by means of suitable valves controlled by the 
governor. 

(c) In very small engines, where economy is not an object, 

governing for temporary changes of load may be effected 
by varying the time of ignition. This is a wasteful 
method. 

In certain cases where both the loads and speeds vary very 
•considerably, as in the case of the locomotive, the regulation is 

135 



136 



DYNAMICS OP MACHINERY 



performed by hand, but in most cases it is accomplished auto- 
matically by an apparatus driven by the motor itself. 

In almost all cases, the direct cause of the action of the governor 
is the variation of speed, while the first result of the variation of 
load is a variation of speed, which in its turn causes the governor 
to act. An exception to this rule may be found, however, in a 
governor at one time employed on the Ball engine, in which the 
variation of load was also a direct cause of the action of the 
governor. 

In almost all governors, use is made of the centrifugal force of 
some rapidly revolving body, counteracted by some other force 
or forces, as gravity, the tension or compression of a spring, the 
resistance of some fluid, friction, the resistance of the mechanism 
operated, etc. 

An exception to the above is to be found in the so-called pressure 
governors, sometimes used on small direct-acting pumps, the air 
pump of the air-brake system, etc., where the pressure in the pump 
itself causes the governor, which is in reality a small auxiliary 
motor, to move the valve of the pump so as to cut off the steam 
supply, and vice versa. 

One of the oldest and most common forms of governor has for 
its fundamental principle the revolving pendulum; hence this will 
be treated first, and its application to governors suitable for service 
will be shown later. 



Simple Revolving Pendulum. 

A heavy body A, whose weight is W, is attached by a weightless 
cord AO at the point 0, and revolves around the vertical axis OB, 
_^ with an angular velocity a, expressed in radians 
per second. Find the height of the pendulum, 
o i.e., the vertical depth of A below 0, which will 
be called h = OB. 

Let AB = r = radius of circle in which the body 
A revolves. 

Then the forces acting on A, and which are bal- 
B anced, are 

1° The weight W represented by Ay. 

W 'cpT 

2° The centrifugal force F = Ax = 

3° The tension of the string, which is equal and 
opposite to A z. 



«, 




y 



Fig. 100. 



Moreover, the similarity of the triangles Azy and OAB gives 

OB _ = Ay = h _ W 
AB yz i 



(Wa 2 r\ > * 






(i) 



GOVERNORS 



137 



Observe that a is expressed in radians per second; also that 
g = 32| feet per second = 386 inches per second. 

Hence to obtain h in feet, write g = 32£, and to obtain h in 
inches write g = 386. 

If N = number of revolutions per minute, we have 

2ttN _ttN 
~ 30 ' 

1 (2) 



a 



h = 



60 
900 g 



ir- 



and 



N = 



7T ▼ 



9 

h 



2> 



(3) 



Observe, also, that for a given speed we have the same value of 
h, whatever the length of string, provided the string is not shorter 

than -^ • 



a' 



If, therefore, we have a set of simple conical pendulums, all 
of which have the same point of attachment 0, all of which revolve 
at the same speed a, and none of whose strings is shorter than 

— 9 . then will these pendulums all revolve in the same horizontal 
or 

plane. 

Example 1. — Given N = 50 revolutions per minute, find h in 
inches. Result: h = 14.08 // . 

Example 2. — Given N = 100 revolutions per minute, find h in 
inches. Result: h = 3.52". 

Example 3. — Given N = 200 revolutions per minute, find h 
in inches. Result: h = 0.88". 



Pendulum Governors. 

Two common forms of pendulum governors are shown in Fig. 101 
and Fig. 102. 

The ordinary pendulum governor 
has a vertical spindle CE which is 
driven by the motor. From this spin- 
dle are hung (at C in Fig. 101, and at 
H and H' respectively in Fig. 102) two 
rods and balls combined (CA and CA' 
in Fig. 101, and HA and H A' respec- 
tively in Fig. 102), which are equal to 
each other in dimensions and weight. 
These balls and rods combined are 
attached to the spindle by pins, and 
revolve with it about its vertical axis. ^. 

From these rods is hung by means of 
two equal and symmetrically located links (ME and M'E in 




138 



DYNAMICS OF MACHINERY 



Fig. 101, and ML and M'U in Fig. 102) the collar LL', which 
is free to slide up and down upon the spindle. While this collar 
rotates about the vertical axis CE, it is connected by suitable 
mechanism with the end of a non-revolving rod (TOQ in Fig. 101, 
and TQS in Fig. 102) which actuates the throttle valve, the cut-off 
gear, or other mechanism that regulates the energy supplied to 
the motor. Evidently, the position occupied by the collar at any 
given instant determines the position, at that instant, of the throttle 
valve, the cut-off gear, etc., and when the position of the collar is 
known the position of these can be determined geometrically. 




Fig. 102. 



Hence the study of the action of the governor practically re- 
duces to a study of the motion of the collar. 

Moreover, when, as shown in Fig. 102, an extra weight R is 
purposely added, and is carried up and down by the collar, on which 
it rests, the governor is called a loaded governor. A study of the 
action of the governor when the load on the motor is constant, 
and when the speed is uniform, is often called the statical treat- 
ment of the problem. 

On the other hand, a study of the behavior of the governor 
during the time that it is adjusting itself to the position suited to 
a new load is called the dynamic treatment of the governor problem. 

Statical Discussion of the Pendulum Governor. 

Assume the motor to be operating under a constant load; then, 
barring periodical variations due to its internal construction, the 
speed is constant, and, therefore, also the position of the collar 
of the governor. 



GOVERNORS 139 

By way of illustration, consider a steam engine, in which the 
governor regulates the cut-off. Then if the load on the engine 
remains constant, the indicator card, and hence the cut-off, the 
position of the collar of the governor, and the angle ACE made 
with the vertical by the center line of the ball and rod combined, 
will also be constant. 

If, therefore, we consider all the forces that act upon one ball 
and rod combined as HA, Fig. 102, and if we take moments about 
H, the sum of the moments of those forces which tend to turn HA 
outwards must be equal in magnitude, and opposite in sense, to 
the sum of the moments of those forces which tend to turn HA 
inwards. The equation which expresses this relation is called the 
moment equation of the governor. 

The forces acting on HA are the following, viz. : 

(a) The centrifugal force of HA. 

(6) The force exerted on HA at M, in consequence of the 
centrifugal force of the link. 

(c) The weight of HA. 

(d) The force exerted on HA at M , in consequence of the weight 

of the link. 

(e) The force exerted on HA at M, in consequence of the resist- 

ance of the valve gear at T, plus the weight of the collar, 
plus that of the extra weight R when this is used. 

The moment equation expresses the fact that the sum of the 
moments of (a) and (b) about H must balance the sum of the mo- 
ments of (c), (d), and (e) about the same point. Moreover, as 
the moment equation is needed for the solution of almost all 
problems connected with governors, it will be necessary to be 
able to deduce it for any special case that may arise in practice. 
In working it out for different cases, the following notation will 
be employed throughout (see Fig. 102) : 

Let Wi = weight of each ball and rod combined. 
W2 = weight of each link. 
P = vertical resistance at the collar, plus the weight of 
the collar, plus extra weight R. 

1 1 = moment of inertia of each ball and rod combined, 

about the axis from which it is suspended, hence 
that of HA about H, or that of H'A' about H '. 

1 2 = moment of inertia of each link about the axis of the 

lower pin, i.e., of ML about L, or of M'L' about L'. 
a = GH = GH'. 
c = KL = KU. 
m = HM= H'M'. 
I = ML= M'L'. 
Xi = distance from H to center of gravity of one ball and 
rod combined, i.e., of HA from H, or of H'A' from 
H'. 



140 DYNAMICS OF MACHINERY 

X2 = distance from L to center of gravity of ML = dis- 
tance from U to center of gravity of M'L' . 
a — angular velocity of the governor per second in 

i = angle ACE = angle A'CE'. 
%' = angle MEC = M'EC. 
N = number of revolutions of governor per minute. 

The relation between i and i' is the following, viz. : 

m sin i + a = I sin i' + c; 

. ., m . . . a — c 
sin v = -y sin t H = 

When 

c = a, and m = I, then ^ = i', a very common case. 

Moreover, let 

Mi= moment about H, of centrifugal force of left-hand ball 
and rod. 

M 2 = moment about H, of force exerted on HA at M, in conse- 
quence of centrifugal force of link ML. 

M 3 = moment about H, of weight of left-hand ball and rod. 

Af 4 = moment about H, of force exerted on HA at M, in con- 
sequence of weight of link ML. 

M b = moment about H, of force exerted on HA at M, in con- 
sequence of P. This is obtained by resolving P into 
two components acting respectively along the two links 
ML and ML', and multiplying the component along 
ML by its leverage about H. The product obtained 
is M 5 . 

In any case of a pendulum governor of this kind, we then 
have 

M l + M 2 = M 3 + M 4 + M 5 (4) 

In order to obtain the moment equation in a form suitable for the 
solution of any given problem, it is necessary to determine and 
substitute the values of these moments in terms of (a) the dimen- 
sions, weights, and moments of inertia of the separate parts of 
the governor, (6) the angle i, or the position of the collar, (c) the 
speed, and (d) the resistance P. In the case of a given governor, 
the dimensions, weights, and moments of inertia of the parts are 
known or can be determined, and hence the moment equation 
expresses the relation between the three unknowns, viz. : (a) the 
speed, (6) the angle i, and (c) the resistance P, any two of which 
being given, the third can be found. Inasmuch as the general 
form is rather long, and as some of the terms are usually small 
when compared with the others, we may often use an approximate 
form where certain ones of the moments are neglected. Thus 
we may, at times, neglect all consideration of the centrifugal force 



GOVERNORS 141 

of, and of the weight of, the links, thus making Mi = M 4 = 0, 
and writing the moment equation M 1 = M 3 + M 5 . Some of the 
simpler cases will be worked out first, and the general case will be 
treated last. 

Case I. 

Assume a governor like that shown in Fig. 1, where a = c = 0. 
Assume also that CM = CM' = ME = M'E, hence that m = I 
and hence that i' = i. Neglect all consideration of the centrif- 
ugal force of, and of the weight of, the links. Then Mi = M 4 = 0. 
The moment equation then becomes 

M ! = M 3 + M b . 

To deduce the expressions for these moments, proceed as follows: 

1° To deduce the value of Mi. 

Assume the weight of the ball and rod combined to be concen- 
trated along the center line CA. 

Let x = distance of any point in the rod and ball from C. 

Let wdx = weight of element of length dx at distance x from 

C, w being a quantity that varies. 
Then we have 



a 2 



(w dx)x sin i = centrifugal force of the element. 



Hence 



a 2 



— (wdx) (x sin i) (x cos i) = moment of centrifugal force 

of element. 

Hence by integration we have for the entire ball and rod 

a 2 C a 2 

Mi = — cos i sin i I wx 2 dx = — Ii cos i sin i. 

9 J 9 

2° To deduce the value of Ms, proceed as follows : 

Wi = weight of ball and rod combined. 
xi= distance from C to center of gravity of ball and rod 
combined. 
Xi sin i = leverage of Wi about C. 
Hence M 3 = WiXi sin i. 

3° To deduce the value of M 5 , proceed as follows, viz. : 

Resolve P into two components along ME and M'E respectively. 
Each of these components is 

^P 

2 cos i 
Hence 

P P 

M h = „ : (CF) = rr : m sin 2 i = Pm sin i. 

2 cos i 2 cos i 



142 DYNAMICS OF MACHINERY 

Therefore for the moment equation we have 



a 2 

— I i cos i sin i = W1X1 sin i + Pm sin L . . . . (5) 
g 

or dividing out by sin i we have 

a 2 

— 1 1 cos 2 = W\Xi + Pra. (6) 

Moreover, since 

a = nf . = -^tt' the moment equation reduces to 
60 30 

ir 2 N 2 

OOQ-hcosi = WiXi + Pm (7) 

Moreover, CE = 2 m cos i (8) 

Examples. 
In the following examples assume 

TTi= 61.94 pounds, 7i= 20,470 (pounds) (inches) 2 , 
I = m = 8.5 inches, #i = 17.93 inches. 
Also use g = 386 inches per second, and observe that it 2 = 9.8697. 

The moment equation then becomes 

0.5816 N 2 cos i = 1110.5842 + 8.5 P (9) 

This may be written in any one of the three following forms, viz.: 

P = 0.06842 N 2 cos i- 130.657 (10) 

N 2 = (1909.5366 + 14.6149 P) sec i (11) 

cos i = (1909.5366 + 14.6149 P)^~ (12) 

Example 1. — Given N = 55 and i = 43°. Find P. Result: 
P = 20.71 pounds. 

Example 2. — Given P = 20.71 pounds and i = 43°. Find N. 
Result: N = 55 r.p.m. 

Example 3. — Given N = 55 and P = 20.71 pounds. Find i. 
Result: i = 43°. 

Example 4. — Given the same data as in 1, 2, and 3. Find CE, 
the depth of the collar below C. Result: CE = 2 (8.5) cos 43° = 
12.43 inches. 

Example 5. — Given JV = 60 and i = 43°. Find P. Result: 
P = 49.483 pounds. 

Example 6. — Given P = 49.483 pounds and i = 43°. Find N. 
Result: N = 60. 

Example 7. — Given AT = 60 and P = 49.483. Find i. Result : 
i = 43°. 

Example 8. — Given the same data as in 5, 6, and 7. Find CE. 
Result: CE = 2(8.5) cos 43° = 12.43 inches. 



GOVERNORS 143 

CASE II. 

Assume a governor like that shown in Fig. 101, where a = c = 0. 

Assume also that CM = CM' = M'E, hence that I = m, and 
hence i' = i. Do not neglect a consideration of the centrifugal 
force of, nor of the weight of, the links. 

The moment equation is, then, as already explained, 

M i + M 2 = M 3 + M 4 + M h . 
To deduce the expressions for these moments, proceed as follows : 
1° For Mi, we have, as in case I, 

a 2 
Mi = — Ii cos i sin i. 

g 

2° To deduce the value of M 2 , proceed as follows, viz. : 

Assume that the weight of the link is concentrated along its 
center line ME. 

Let x = distance from E of any point in the link. 

Let wdx = weight of element of length dx, at distance x from 
E. Then we have 

a 2 

— (w dx) (x sin i) = centrifugal force of element. 

Q 

Resolve this into two parallel components at M and E respectively. 

Then we have 

a 2 x 

component at M is — (w dx) (x sin i) — 
g m 

The component at E has no effect on the ball and rod, and is 
balanced by the corresponding component from the link on the 
opposite side of the spindle, which is equal and opposite to it. 
Hence we have moment of component at M, about C, is 

— (w dx) (x sin i) —m cos i = — cos i sin ix 2 w dx. 

g m g 

Hence by integration we obtain for the moment about C, due to 
the centrifugal force of the link, 

a 2 I or 

Mi = — cos i sin i I wx 2 dx = — 1% cos i sin i. 

g J g 

3° For M 3 , we have, as in case I, M 3 = W&i sin i. 
4° To deduce the value of ilf 4 , proceed as follows, viz. : 

To 

Let W m = weight of upper end of link = W2 — • 

L e t W\ = weight of lower end of link = W 2 

m 

Hence, moment of weight of upper end about C is 

W 2 — m sin i = W2X2 sin i. 
m 



144 DYNAMICS OF MACHINERY 

Now resolve the weight of the lower end into two components, one 
along the link, and the other horizontal; we have that the second 
produces no effect on the ball and rod, and is balanced by the 
corresponding component of the weight of the lower end of the 
other link. The component along the link, on the other hand, is 

TT , m — x 2 

W 2 :' 

m cos^ 
The moment of this component about C is 

W r m sin 2 i =2 W 2 (m — £2) sin i. 

m cos ^ 

Hence we have 

Mi = W 2 x 2 cos i + 2 W 2 (m — x 2 ) sin i, 
or 

Mi = W 2 (x 2 + 2 m — 2 x 2 ) sin i = W 2 (2 m — x 2 ) sin i. 

5° For M 5 we have, as in case I, 

M 5 = Pm sin i. 

Hence the moment equation Mi + M 2 = M s + Mi + Ms becomes 
in this case 

— 1 1 cos i sin i H 1 2 cos t sin i = W1X1 sin i + W 2 (2 m — # 2 ) sin i 

9 9 

-{-Pm sin ^ (13) 

Dividing out by sin i, and simplifying, we have 

(h + U) cos i = TTiXi + W 2 (2m- x 2 ) + Pm. . (14) 



a 2 



9 

ttN 



Moreover, since a. = — the moment equation reduces to 

oU 



ttW 2 



■ H (/1 + /«) cos i = TTxXi + TF 2 (2 m - s 2 ) + Pm. (15) 

Usually the link is a symmetrical body, and hence X2 = -^ 
In this case, therefore, the moment equation reduces to 

gjjjpCTi + 1 2) cos i = TT^i + 1 TF 2 m + Pm. . (16) 
Moreover, CE = 2 m sin t (17) 

Examples. 

In the following examples, assume the same data as in case I, 
viz.: 

Wi = 61.94 pounds, h = 20,470 pounds (inches) 2 , I = m = 
8.5 inches, #i = 17.93 inches, # = 386 inches per second, and 



GOVERNORS 145 

in addition the following, viz.: W 2 = 1.168 pounds, I 2 = 16.5 
pounds (inches) 2 , x 2 = - (8.5) = 4.25 inches. 
The moment equation then becomes 

0.5816 N 2 cos i = 1125.4762 + 8.5 P. . . . (18) 
This may be written in any one of the three following forms : 

P = 0.06842 N 2 cos i - 132.409 (19) 

N 2 = (1935.1379 + 14.6149 P) sec i. ... (20) 

cos i = (1935.1379 + 14.6149 P) ~ (21) 

Example 1. —Given N = 55 and i = 43°. Find P. Result: 
P = 18.958 pounds. 

Example 2. — Given P = 18.958 pounds and i = 43°. Find N. 
Result: N = 55. 

Example 3. — Given N = 55 and P = 18.958. Find i. Result: 
i = 43°. 

Example 4. — Given the same data as in 1, 2, and 3. Find CE. 
Result: CE = 2 (8.5) cos 43° = 12.43 inches. 

Example 5. — Given N = 60 and i = 43°. Find P. Result: 
P = 47.731 pounds.- 

Example 6. — Given P = 47.731 pounds and i = 43°. Find N. 
Result: N = 60. 

Example 7. — Given N = 60 and P = 47.731. Find i Result: 
i = 43°. 

Example 8. — Given the same data as in 5, 6, and 7. Find CE. 
Result CE = 2 (8.5) cos 43° = 12.43 inches. 

Graphical Representation of the Moment Equation Applied to Equa- 
tion (5). 

It is often desirable to represent the moment equation of 
a governor graphically. Such a representation will be explained 
in the case of equation (5) , page c 

142, which is the moment equa- y . /f\\ 

tion for case I. c | j I 1 1 

In the figure, lay off as ab- tfM'fH^ 

scissse the horizontal distances \ \jtf p 

of the center of gravity of the p i I I M 

ball and rod from the axis of -|S u.s^ ir iSJ^ 

the spindle. Calling these ab- F - 1Q3 

scissae x, we have x = X\ sin i, 

where X\ = distance of center of gravity of ball and rod from 

C, hence x x is a constant and x is proportional to sin i. If we 

plot on the diagram a series of values of x, we can obtain for 



146 DYNAMICS OF MACHINERY 

each the corresponding value of the angle i, from the equation 

X 

sin i = — - From each of these values of i compute, and lay off 

X\ 

as ordinates, (a) the corresponding values of Pm sin i = x, 

#1 
thus obtaining the curve marked PP, in the figure (when P 
varies with x this is a curve, but when P is constant for all values 
of # it is a straight line passing through 0) ; (b) the corresponding 
values of W\X\ sin i = WiX, the moment, about C, of the weight of 
the ball and rod, and add them to the corresponding ordinates of 
the curve marked PP in the figure. The result is the curve marked 
CC in the figure. Its ordinates will, in virtue of equation (5), page 
142, be equal to those representing the corresponding moments, 
about C, of the centrifugal force of ball and rod. From these, and 

a 2 
the fact that Mi = — I\ cos i sin i, we can compute the correspond- 

ing values of a, and thus plot the curve marked aa, whose ordi- 
nates represent the speeds of the spindle in radians per second. 

We are then in a position to predict the speed corresponding 
to any value of a:,. hence that corresponding to any value of the 
angle i, hence that corresponding to any position of the collar, 
and hence that corresponding to any cut-off. 

Some Rough Approximations. 

Some very inexact approximations will now be deduced. While 
they are liable to give results as much as fifteen per cent in error, 
nevertheless they are sometimes employed for the purpose of 
making a preliminary calculation, the results of which have to be 
altered by means of more exact methods. 

Let B = weight of each ball, assumed concentrated at its 
center. 

R = weight of each rod (CN in Fig. 101), assumed concen- 
trated along its center line. 

b = CA (Fig. 101) = distance from point of suspension C 
to center of ball. 

r = radius of ball. 

X = b — r = CN = length of rod. 

We then have the following equations, part of which are approx- 
imate: 

7, =BV + ^ (22) 

/, = ^p (23) 

Wi = B + R (24) 



GOVERNORS 147 

*' = — fFT~ (25) 

*»-5 (26) 

If we substitute these values in equation (7), page 142, of case I, 
where the centrifugal force of, and the weight of, the link is dis- 
regarded, we have 

— - - j £6 2 + -y- > cos i = Bb + — + Pra, . . (27) 

while equation (25) gives 

Bb + f 

Xl= B + R (28) 

If, now, as a further approximation, we disregard also the rods, 
this being equivalent to writing R = 0, these equations become 

%^-Bb 2 cos i = Bb + Pm, (29) 

900 </ v ' 

xi=b (30) 

These equations, as has been stated, are very inexact, as will be 
illustrated by the following examples. 

Example 1. — From equation (29) find the weight B, which, 
concentrated at the center of gravity of the ball and rod, would 
overcome the resistance P = 18.958 pounds, found in case II for 
55 r.p.m.; i.e., given P = 18.958, m = 8.5 inches, b = 17.93 inches, 
and i = 43°, find B. Result: B = 70.8 pounds. This is very 
considerably larger than 61.94 pounds, the total weight of ball and 
rod given in cases I and II. 

Example 2. — From equation (29) find the weight B which, con- 
centrated at the center of gravity of the ball and rod, would over- 
come the resistance P = 47.731 pounds, found in case II for 60 
r.p.m.; i.e., given P = 47.731 pounds, m = 8.5 inches, b = 17.93 
inches, and i = 43°. Result: B = 68.2. This is considerably 
larger than the total weight, 61.94 pounds of ball and rod, given in 
cases I and II. 

General Discussion of the Pendulum Governor. 

Having thus far discussed certain special cases, for the sake 
of simplicity, and for the purpose of illustration by means of 
examples that do not involve much complexity, we will now pro- 
ceed to the deduction of the moment equation for the general 
case. 

Fig. 102 shows the usual construction of the pendulum governor. 



148 DYNAMICS OF MACHINERY 

As already explained (page 140), if we consider all the forces 
acting on one ball and rod we shall have the equation 

Mi + M 2 = Ms + M 4 + M 5j (1) 

where 

M i = moment, about H, of centrifugal force of left-hand ball 
and rod. 

Mi = moment, about H, of force exerted on HA at M, in con- 
sequence of the centrifugal force of the link ML. 

Ms = moment, about H, of the weight of left-hand ball and rod. 

M± = moment, about H, of the force exerted on HA at M, in 
consequence of the weight of the link ML. 

Ms = moment, about H, of force exerted on HA at M, in conse- 
quence of the force P which includes (a) the weight of 
the collar, (b) the weight of the extra load on the collar, 
and (c) the force at the collar required to overcome the 
resistance due to the valve gear and the internal fric- 
tion of the governor. 

Moreover, we have also the following equation connecting i' and 
i which has already been deduced, viz.: 

. ., m . a — c t . 

sini = y sm i -\ = — (2) 

We will now determine the values of the several terms in equa- 
tion (1). 

1° To deduce the value of Mi. 

For this purpose, we will assume the weight of the ball and rod 
to be concentrated along the center line. 

Let x = distance of any point in the rod and ball from H. 
Let wdx = weight of element of length dx at distance x from H. 

Then we have 

a 2 

— (w dx) (a + x sin i) = centrifugal force of the element. 

9 

a 2 

— (w dx) (a + x sin i) (x cos i) = moment of centrifugal force 

of element. 

Hence 

Mi = — < ( / wx 2 dx Jcos i sin i + a I I wx dx J cos i > 

or 

a 2 ( ) 

Mi = — < 1 1 cos i sin i + aW \xi cos i \ . 

9 ( ) 

2° To deduce the value of M 2 . 

For this purpose we will assume that the weight of the link 
is concentrated along its center line ML. 



GOVERNORS 149 

Let x — distance from L to any point in the link. 

Let wdx = weight of element of length dx at distance x from L. 

Then we have 

a 2 

— (w dx) (c + x sin i') = centrifugal force of element. 
9 

Resolve this into parallel components at M and L respectively. 
Then 

— (w dx) (c + x sin i') j = component at M . 

The component at L has no effect on the ball and rod, the corre- 
sponding component from the link on the opposite side being equal 
and opposite. 

Hence 

a 2 x 

— (w dx) (c + x sin i r ) j m cos i = moment, about H, of com- 

^ ponent at Mi. 

Hence 

Gt 2 7YI COS % [ / f* \ / I 

M 2 = j j ( I wx 2 dx) sin i' + c ( / wa; c?x 

or 

_ T a 2 m cos ii T . ., . 1TT ( 

M 2 = j — 1 1 2 sin i' + cW 2 x 2 \ • 

9 I I ) 

3° For M 3 we have 

Mz = WiXi sin i. 

4° Let W m = weight of upper end of link. /. W m = W 2 y • 

Z — x 2 
Let TFi = weight of lower end of link. /. Wi = W 2 



I 



To 

Hence moment of W m about H is W 2 -j-m sin i. 



Now resolve TF^ into two components, one along the link, and 
the other horizontal, the latter component having no effect on 
the ball and rod. 

I — x 2 
The component along the link is W 2 j r, > and its moment 

L COS % 

about H is 

W 2 7 ~ X * m sin (i + i') = W 2 — r~ 2 m (sin i + cos i tan i f ) . 
I cos %' I 

Hence it follows that 

M A = W 2 ^jm sin i + W 2 , 2 m (sin t + cos i tan t'). 



150 DYNAMICS OF MACHINERY 

5° To deduce the value of M 5 we proceed as follows : 
Resolve the vertical force P into two components along ML 
and M'L' respectively. Each of these components is 

P 
2 cos %' 

Hence 

P Pm 

Ms = ~ t. m sin (i -+- %') = —p-r (sin i + cos i tan i') . 

2 cos i . 2 

Hence, substituting these values in equation (1), we obtain for the 
moment equation 

C? \ 771 C771 # 

— X 1 1 cos i sin i + aW^ cos z -f- -7- 1 2 cos i sin i' + -7- TF 2 £ 2 cos 2 > 
( I I - ) 

= TF1X1 sin i + TF2 y w sin i + TT 2 — ^ — - m (sin i -f cos z tan i') 

- -^- (sin i -f cos i tan t') (3) 

In employing the moment equation to study the action of the 
governor, it is better to have it in such a form that it contains only 
quantities that are directly measured or weighed, or observed. 

Thus if W a = weight of ball and rod at A, and HA = b, then 
TTiXi = bW a . 

Hence, substitute bW a for W\Xi. 

Substitute W m iovW2j' 

Substitute Wi for W 2 l - Z j^ 3 ■ 

CL — C 

Substitute sin i -\ j — for sin i' in the third term. 

Making these substitutions, we obtain 

CX? V 771? I (a — c) 771 1 ~Y 

— \ I\ + -p h cos i sin i + abW a + - — 7^ — 7 2 + cmW m cos i > 

— (&TF + TTiWm) sin ^ — mTFz (sin i + cos i tan i') 

= — — (sin i -f- cos z tan i') (4) 

If t\ = time of a single vibration of ball and rod when swinging 
about H, and U = time of a single vibration of the link when swing- 
ing about L, we shall have 

I^^bWa and h = t z£lW m ;. . . . (5> 

7T 7T 



GOVERNORS 151 

then, if these substitutions be made in (4), we shall have as a 
result a form that contains only quantities that have been directly 
measured, weighed, or observed. 

Value of GK. 

In designing a governor for a certain service, or in endeavoring 
to predict the behavior of a governor which is already constructed, 
it is necessary to study the relation between 1°, the position of the 
collar; 2°, the speed of the spindle; and 3°, the value of P. 

Inasmuch, however, as the moment equation does not contain 
GK directly but does contain the angle i, we need to know the 
relations between GK and i, and, incidentally, between i' and i, 
so that when GK is known i can be found, and vice versa. 

The equations which express these relations, as will be evident 
from Fig. 102, are 

GK = m sin i + I sin i' (6) 

.. m . . . a — c ,_. 

sin i = -r sin i -\ j — , . (7) 

In this case graphical solutions will be found convenient; thus we 
may plot one curve having values of i for abscissae and values of 
i', as determined from equation (7), for ordinates, and another curve 
having values of i for abscissae, and values of GK, as determined 
from equation (6), for ordinates. 

General Discussion. 

In discussing the action of any given pendulum governor of the 
kind shown in Fig. 102, we start with the three equations: 

. ., m . a — c m 

sini =ysiniH j — (1; 

— ) Ui + -p- 1 2 cos i sin i + abW a H ™ li + cmW m cos i > 

— {bW a + mW m ) sin i — mWi (sin i + cos i tan i') 

= —=- (sin i + cos i tan i') (2) 

GK = m cos i + I cos i' (3) 

The value of i' can be determined in terms of i from equation 
(1), and substituted in equation (2); hence we may say that we 
have in equation (2) the following three variables, viz., i, n, and 
P. Now if any two of these are known the third can be determined, 
and, on the other hand, if only one is known we can obtain an 
equation expressing the relation between the other two; or this 
relation can be represented graphically by a plane curve. 



152 DYNAMICS OF MACHINERY 

As to GK, that is known as soon as i is known. Hence we may 
have the following three cases: 

1° Given i or GK, to determine the relation between P and a, or, 

which amounts to the same, between P and n. 
2° Given a or n, to determine the relation between P and t, 

or, which amounts to the same, between P and GK. 
3° Given P, to determine the relation between i and a, or, 

which amounts to the same, between GK and a, or between 

GK and n, or between i and n. 

Either one of these three relations may be used in studying 
the action of any proposed governor, or of any governor already 
constructed. 

Of course, in order to obtain the relations stated, we need to 
know the quantities 

Ji, h, Wi, W 2 , x h x 2 , a, c, I, m. 

In the case of a proposed governor, these are to be obtained by 
calculation from the proposed dimensions. In the case of a gov- 
ernor already constructed, they are to be obtained as follows, viz. 

The last four are of course obtained by measurement; the third 
and fourth by weighing; the fifth and sixth by weighing the corre- 
sponding parts at each end; and the first two by supporting them 
on knife-edges and letting them oscillate by gravity, and counting 
the number of oscillations per minute. 

When the motor is running at its normal speed, and the supply 
of steam or water is just adapted to the work to be done, the gover- 
nor will probably have to exert a certain pull p at the collar to 
keep the mechanism in the proper place, and this, exclusive of 
any pull due to friction. 

Now if the speed increases, the collar does not move until it has 
increased so much as to develop an additional pull sufficient to 
overcome the friction F, and if the speed decreases, the collar will 
not move until it has decreased so much as to diminish the pull 
by an amount equal to the friction /. 

Hence it follows that, if the collar stands at a certain height 
GK = ho, the corresponding value of i being i , and if, from the 
relation between P and n corresponding to this value of i, we 
determine the value of n = n corresponding to P = po, where 
the latter is the pull required to keep the mechanism in place with 
no friction operating either way, then if F and / are the values 
of F and / corresponding to % = i , and if we determine from the 
above-stated relation between P and n the value n = n\, corre- 
sponding to P = po + Fq, and the value n = n 2 , corresponding 
to P = po — fo, then the speed of the governor may vary anywhere 
between n = n\ and n = n 2 without changing the position of the 
collar, and hence without causing the governor to act. 

Of course it is desirable for good working of the motor to have 
n\ — no and n Q — n 2 , and hence n\ — n 2 , as small as possible; and 



GOVERNORS 153 

if it is too large for good work, some change should be made in 
the governor or in the mechanism connected with it, so as to secure 
a better regulation. 

Now let us suppose that we have reduced the quantity n x — n 2 
to sufficiently small proportions so that the governor does regulate 
quickly; then the next question that requires to be investigated is 
the following, viz. : 

It is plain that the position of the regulating mechanism (as, for 
instance, the position of the cut-off gear of a steam engine) is differ- 
ent for every different amount of energy, or amount of load (as, 
for instance, for every different cut-off). 

Hence, let us consider the maximum and the minimum load on 
the engine for which a good degree of regulation is desired, and 
determine the positions of the collar, or the values of i correspond- 
ing to these two loads respectively. 

Let these values of i be i = i max . and i = i m \n.- Then deter- 
mine the relations between P and n corresponding to each of these 
two angles. Then in the equation between P and n corresponding 
to i max . substitute for P the value of p + F corresponding to this 
value of i y and thus determine n max .. Then in the equation corre- 
sponding to i m i n . substitute for p the value of p — / corresponding 
to this value of i, and thus determine n m i n .. Now the quantity 
n max . — n m i n . should of course be small, in order to secure good 
regulation. 

Hence it is plain that a governor does not regulate well unless 
we have both 
small n\ — n 2 and n m&x . — n m i n . 

On the other hand, we must guard against another defect which 
is liable to arise when these quantities are small, and it is that 
when the limiting speed n\ or n 2 has been reached, and the gov- 
ernor begins to regulate, inertia may carry the collar beyond the 
point where the energy supplied is just suitable for the load; and 
hence the change of speed, being carried beyond the proper point, 
it has to return, and thus are produced a series of variations from 
too high to too low a speed, and vice versa, which gives a very 
irregular motion, called racing or hunting. 

Of course increasing the friction tends to prevent racing, as also 
the introduction of a dashpot; but thus the governor is made 
more sluggish in its action, and therefore the designer of a governor 
always aims to make the quantities 

rii — n 2 and n max , — n m i n . 

as small as possible consistent with the absence of racing or hunt- 
ing. 

Of course in order to make the study complete we need to know 
the values of p, F, and /, and how they vary for every different 
position of the collar. For this purpose experiment is needed, 
and, moreover, these quantities depend, not on the governor only, 



154 



DYNAMICS OF MACHINERY 



but also upon the regulating mechanism which is moved by the 
governor. 

Example. — Given a pendulum governor having the following 
constants, the units being pounds, inches, and seconds : 

TFi = 61.94, h= 20,470, m = 8.5, 

W 2 = 1.168, h = 16.5, I = 6.5, 

W m = 0.635, a = 0, x 1 = 17.933, 

W i = 0.533, c = 1.875, g = 386. 

Let JV = number of revolutions of the spindle per minute. 
Find relations between P and N corresponding to the following 
values of angle i : 

40°, 41°, 42°, 43°, 44°, 45°, 46°, 47°, 48°, 49°, 50°. 

Also find corresponding values of h. 



Solution. 



sin 2/ = 



msmi 



I 



h = m cos i + I cos %'. 




cos i sin i + « 



abW a + 



(a — c) m 
T 2 



I 2 + cm W m cos i 






— (bWa + mW m ) sin i — mWi (sin i + cos i tan i r ) 

Pm, ... .,. 
= -^-(sini + cost tan i ), 



(4) 



which after the substitution of the data reduces to 

' " (10,253) sin 2 i + ^-^- (2.2711) cos % I N 2 



(900g^ , ~^ / " ' 900 g 
— (1116.17) smi — 4.5305 (sini + cos i tan i') = 4.25 (sini 
+ cos i tan i') P (5) 

By means of (9), coupled with (7), we can deduce the equations 
that give the relations between P and N for the following values 
of i, viz. : 

40°, 41°, 42°, 43°, 44°, 45°, 46°, 47°, 48°, 49°, and 50°. 

Only the results will be given here. 







Value of 


Value of 






1. 


Value of P. 


N, when 
P=0. 


P, when 
N=0. 


GK. 


Diff. 


40° 
41° 


P= .058699 N 2 - 147.85 
P= .057574 N 2 - 147. 20 


50.19 
50.56 


-147.85 
-147.20 


11.930 
11.758 




.172 


42° 


P= .056464 N 2 - 146. 61 


50.96 


-146.61 


11.581 


.177 


43° 


P=. 055354 N 2 - 146. 05 


51.37 


-146.05 


11.400 


.181 


44° 


P= .054239 N 2 - 145. 50 


51.79 


-145.50 


11.214 


.186 


45° 


P= .053136 N 2 - 145. 01 


52.24 


-145.01 


11.025 


.189 


46° 


P=. 052023 N 2 - 144. 52 


52.71 


-144.52 


10.831 


.194 


47° 


P= .050896 N 2 - 144. 02 


53.20 


-144.02 


10.634 


.197 


48° 


P= .049784 N 2 - 143. 59 


53.71 


-143.59 


10.433 


.201 


49° 


P =. 048661 N 2 - 143. 22 


54.25 


-143.22 


10.229 


.204 


50° 


P=. 047538 N 2 - 142. 73 


54.79 


-142.73 


10.020 


.209 



GOVERNORS 



155 



Examples. 

This set of examples refers to the governor for which the above 
equations have been deduced. 



Example 1. — Given i = 43 c 
Result: P = 21.395 pounds. 

Example 2. — Given i = 43' 
Result: P = 53.324 pounds. 

Example 3. — Given i = 40° 
Result: P = 17.006 pounds. 

Example 4. — Given i = 40 c 
Result: P = 63.466. 



Find P when N = 55 r.p.m. 

Find P when iV = 60 r.p.m. 

Find P when N = 53 r.p.m. 

Find P when iV = 60 r.p.m. 



As would naturally be expected, for a given angle i, the value 
of P corresponding to a greater speed is greater, and for the same 
speed of spindle, a larger value of P. is accompanied by a smaller 
value of the angle i, and hence of GK. 



Loaded and Unloaded Pendulum Governor. 

The distinction between a loaded and an unloaded pendulum 
governor is the following, viz. : 

In the case of a governor like that shown in Fig. 101, the value 
of P includes 

(a) The force required at the collar, to hold the regulating 
mechanism in place, including, of course, the weight of any of the 
regulating mechanism that hangs from the collar. 

(b) The resistance of friction. 

(c) The force at the collar required to overcome any resistance 
that may be due to the dashpot, when there is one, and 

(d) The weight of the collar itself. 

When P includes nothing in addition to the above, the governor 
is said to be unloaded. When, on 
the other hand, an additional weight 
is placed on the collar, as is indicated 
in Fig. 102, the governor is said to be 
loaded. 

In making a qualitative study of 
the difference in behavior between 
an unloaded and a loaded pendulum 
governor, it will be sufficient for the 
first part to consider the case of a 
governor like that shown in the 
figure, in which the links are attached 
to the balls by forks, and where 
CA = CA' = AE = A'E, and conse- 
quently where i' — i, and to dis- 
regard the centrifugal force and the weight of the links, and also 
of the rods. 




Fig. 104. 



156 DYNAMICS OF MACHINERY 

Let a = speed of spindle in radians per second. 
N = speed of spindle in revolutions per minute. 
g = 386 inches per second. 
B = weight of each ball in pounds. 
b = CA = CA' = AE = A'E in inches. 
P have the same meaning as heretofore. Evidently its 
value in the case of a loaded governor is greater than it 
is in the case of the same governor unloaded. 
Pi = value of P in the case of an unloaded governor. 
Pi = value of P in the case of the same governor loaded. 
Then the load is evidently equal to Pi — P 2 . 
With this notation, and with these assumptions, the moment 
equation (see equation (5), page 142) becomes 

a 2 

— Bb 2 cos i sin i = (B + P) b sin i (1) 

Q 

Dividing through by b sin i, and substituting a = — -, it reduces to 

g^BftcoBt-B+P, (2) 

and since CE = 2b cos i = depth of collar below the point of 
suspension C, the moment equation may be reduced to either of 
the following forms, viz. : 

CB=^(l + g. . ..... (3) 

^,70,380/, , P 



( 1 + b) < 4 > 



CE 

From equations (3) and (4) we may draw the following conclu- 
sions, viz. : 

(a) For a given speed of spindle, the larger P, the larger is CE. 

(b) For a given speed of spindle, the value of CE is larger in 
the case of a loaded governor than it is in the case of the same 
governor unloaded. 

(c) For a given value of CE, the larger P, the greater is the speed 
of spindle. 

(d) For a given value of CE, the speed of spindle is greater in the 
case of a loaded governor than it is in the case of the same governor 
unloaded. 

(e) If an unloaded governor, and the same governor loaded, 
are to have the values of CE equal to each other, when N 2 = speed 
of spindle of unloaded in revolutions per minute, and Ni = speed 
of spindle of loaded in revolutions per minute, then, in virtue of 
equation (4), we have 





GOVERNORS 157 



Hence, if a given engine, provided with this unloaded governor, 
have the speed of its main shaft equal to n in revolutions per 
minute, and if the governor is so geared that N 2 = rn, then if we 
add a load on the collar whose magnitude is Pi — P 2 we shall have 
to gear the governor up so that 



JVi =r 




n, 



and, when this is done, both governors will always have the same 
value of CE for the same speed of the engine, and hence for the 
same cut-off, and the same variations of speed will correspond to 
the same variations of cut-off. 

(/) If, on the other hand, we make no change in the gearing up, 
and hence if the speed of spindle is the same in the two, then we 
have Ni = N 2 = rn, we shall then have, that the values of CE will 
not be the same in the two cases, but will bear a certain ratio to each 
other which may be deduced from equation (3) as follows, viz. : 

In virtue of equation (3), we shall have 



L/2E2 — 



2,,y-t 2 



r'n 



_ 70,380 /P, 
ClBl " ~FW" i 1 + B 

Hence , Pi 

C 1 E 1 = _ 1 + ~B 
C2E2 n , P2 

1+ p 

In order to use the loaded governor on the engine, we shall have 
to move the point of suspension higher on the spindle, and we 
may have to lengthen the rods. The height on the spindle to 
which the point C will have to be raised is such that for some one 
cut-off, preferably one corresponding to the mean speed, the collar 
shall be in the right place. When this is done we shall have 1°; 
the variations of motion for the same change of speed will be 
greater for the loaded than for the unloaded governor, and hence 
the change of speed corresponding to a given change of cut-off 
will be less in the case of the loaded than in that of the unloaded 
governor. 

Change of Speed Required to Overcome a Given Resistance at the 
Collar in a Loaded Governor, as Compared with that Required 
in the Case of the Same Governor Unloaded. 

Instead of making the assumptions and approximations employed 
in the first part of this study, we will consider the case of the actual 
governor, to which apply the equations on page 154, which express 



158 DYNAMICS OF MACHINERY 

the relations between P and N for various values of the angle i, 
and will develop the method of procedure by means of a numerical 
example. 

Assume the angle i = 43°, and the normal speed of the spindle 
when the governor is unloaded to be N = 55 r.p.m. 

For i = 43° we have P = 0.055354 N 2 - 146.05. 

When N = 55, this gives P = 21.396 pounds. 

Hence we will assume that, when the governor is unloaded the 
value of P is P 2 = 21.396 pounds, which includes the weight of 
the collar and the force required to hold in place the regulating 
mechanism when the engine is running at a constant load corre- 
sponding to such a speed as will cause the speed of the governor 
spindle to be N = 55 r.p.m. 

If now the speed be increased to 56 r.p.m., which is an increase 
of 1.82 per cent, and hence, if the speed of the engine increases by 
1.82 per cent, we have P= 27.840, and since this exceeds 21.396 by 
6.144, it follows that if the extra resistance at the collar when the 
governor starts to lift is 6.144 pounds, it would require a change of 
speed of the engine of 1.82 per cent before the governor would 
lift. 

We will now compare with the above the action that the gover- 
nor would exert were we to load it, and, at the same time, to 
gear it up to a speed corresponding to the load. Suppose the load 
to be such as will require a gearing up that will cause the speed 
of the spindle to be 70 r.p.m. instead of 55 r.p.m., both for 
the same speed of engine. Then if we substitute N = 70 in the 
equation for P, we obtain P = 125.185 pounds; and since 125.185 
— 21.396 = 103.789 pounds. Hence the load to be added on 
the collar to render the normal speed of spindle 70 r.p.m. is 103.789 
pounds. 

Now, suppose the speed of spindle to become 70.8 r.p.m. instead 
of 70 r.p.m., i.e., to undergo an increase of 1.14 per cent, let us 
find what extra resistance the governor will overcome. The 
equation for P gives us P = 125.185 pounds when N = 70, and 
P = 131.419 pounds when N = 70.8. Hence the extra resistance 
which the governor will overcome is 131.419 — 125.185 = 6.234 
pounds. Consequently it appears that if, when unloaded and 
with the speed of spindle 55 r.p.m., the extra resistance is 6.144 
pounds the speed of the engine must increase by 1.82 per cent 
before the governor lifts, whereas when a load of 103.789 pounds 
is added on the collar, and the governor spindle is geared up so 
that it runs at 70 r.p.m. for the same speed of engine for which 
it ran at 55 when unloaded, then the governor will lift before 
the engine has increased its speed by 1.14 per cent. The facts 
developed by this illustration are often expressed by saying 
that a loaded governor has more power than the same governor 
unloaded. 



GOVERNORS 



159 



Loaded Governor on Engines with High Rotative Speed. 

When applying a pendulum governor to an engine with high 
rotative speed, the use of an unloaded governor would often 
involve running its spindle at a speed so much below that of the 
main shaft of the engine that the reduction of speed required 
might in certain cases be considered undesirable. In such cases 
a loaded governor could be used. 



Approximate Limits of Variation of Speed. 

The manner of making a study of the limits of variation of speed, 
between the longest and the shortest cut-offs, and its relation to 
the corresponding travel of the collar, 
has already been explained in the 
general discussion, and more or less 
considered in other propositions. The 
following approximate method, how- 
ever, will be given; for, while it is an 
inexact method, nevertheless some- 
thing of this kind is often given, and 
employed to obtain an approximate 
result. For this purpose we shall con- 
sider the case where the links are 
attached to the balls by forks (see 
Fig. 105), where CA = CA' = AE = . 
A'E, and where consequently i' =i. 
We shall also disregard the centrifugal 
force of, and the weight of, the links 
and rods. 

The moment equation then becomes 




Fig. 105. 



a 



— Bb 2 cos i sin i = (B + P)b sin i, 

g 



which may be written 



CE= 2 b cost = 



2<7 



a' 



i+ 5 J, 



(i) 



(2) 



where a = speed of spindle in radians per second and CE = depth 
of collar below C, the point of suspension, corresponding to the 
speed a. 

In what follows, we will let a denote the mean speed in radians 
per second, and let 2 h denote the corresponding value of CE. 
We then have 

Suppose now that we require that the extreme variation of speed 



160 DYNAMICS OF MACHINERY 

shall not be more than — th of the mean speed a. Then we have 

m 

Least speed = all — - — ) and Greatest speed = all +- — )• 
\ 2 m) r \ 2 m) 

The least speed corresponds to the greatest value of CE, which 
will be denoted by 2 h + K\. 

The greatest speed corresponds to the least value of CE, which 
will be denoted by 2 h — K 2 . 

Then if we denote by K the total vertical movement of the collar, 
we shall have 

K=K 1 + K 2 (3) 

Hence, from equation (2), by making the suitable substitutions, 
we have 

2h+K ^ j 1 2 !_iy ( i+ i)-H i -^T- (4) 

V 2 m) 

.\ K 1 = 2h\(l-^Y-ll=2h\± + ^ 2 +^ 3 +etc.{ (5) 
( \ 2 m) ) (m 4 m 2 2 m 6 ) 

and 2h-K,= 2 % (l +g - 2h(l +^f . (6) 

« ( 1 + o 

\ 2 m) 

:. X 2 =2^jl-(l+^yU2/ i U- z ^ + ^- 3 etc.[. (7) 
( \ 2 m) \ I m 4m 2 2 m 3 N 



Hence follows K= K 1 -\- K 



\2 



= 2h)- + -\ + etc.l, ... (8) 
(mm 6 ) 



or, omitting powers of — above the first, we obtain, 

m 



K = 2h (l)> < 9 > 



The first member is the ratio of the total travel of the collar to 
the value of CE corresponding to the mean speed a. 

Isochronism. 

In the case of the ordinary pendulum governor, when the collar 
is in its highest position, and when, consequently, CE (Fig. 101) 
has its least value, we shall find that (a) the cut-off is the shortest, 
and (b) the corresponding speed of the spindle, and hence also 
that of the engine, will have their greatest values. 



GOVERNORS 161 

On the other hand, when the collar is in its lowest position, and 
when, consequently, CE (Fig. 101) has its greatest value, we shall 
find that (a) the cut-off is the longest, and (6) the corresponding 
speed of spindle, and hence also that of the engine, will have their 
least values. For every different position of the collar, and hence 
for every different cut-off, there is a definite speed of spindle, which 
can be determined from the moment equation, which will result 
in the existence of equilibrium between the forces acting on the 
ball and rod. Moreover, this definite speed is different for every 
different position of the collar. 

An isochronous governor is one in which, whatever the cut-off, 
there is only one speed of spindle, and hence only one speed of the 
engine, which will result in the existence of equilibrium between 
the forces acting on the ball and rod; this being the special speed 
for which the governor was designed and constructed. When 
the engine runs at any other speed, equilibrium between the forces 
acting on the ball and rod ceases to exist, and, as a result, the 
balls move either outwards or inwards and continue to do so until 
the original speed is again attained. 

To explain the action of an isochronous governor, assume an 
engine which it controls running under a constant load and a 
constant steam pressure, and at the definite speed for which the 
governor was designed and constructed. As long as the load and 
steam pressure remain constant there will be no tendency for the 
collar to rise or fall, and no tendency for the balls to move either 
outwards or inwards. 

Next, suppose that, either through change of load, or change of 
steam pressure, the speed of the engine, and hence that of the 
spindle, changes. Then will the equilibrium between the forces 
acting on the ball and rod be destroyed, and hence the balls will 
move, outwards for an increase, and inwards for a decrease, of 
speed; hence the collar will change its position, and the cut-off 
will change, and these will continue to change, a part of the time 
in one direction, and a part of the time in the opposite direction, 
until the speed has again reached a value equal to the original 
speed, and, at the same time, the cut-off has attained such a value 
that the energy exerted by the engine is just suited to balance the 
load on the engine, at the original speed. 

When, and if, such a condition occurs, the engine will go on 
running at its original speed, under its new load, and there will 
again be equilibrium between the forces acting on the ball and rod. 

The difficulty with such a governor is that, when the speed 
changes, the inertia of the balls carries them too far, and, when 
moving back, they go too far again; or, in other words, such gov- 
ernors are too sensitive, and keep changing the position of the 
regulating mechanism by too great amounts, as the balls will 
always be in motion, outwards or inwards, whenever the engine 
is running at any speed which differs from the original speed for 
which the governor was designed and constructed. 



162 DYNAMICS OF MACHINERY 

This oscillating motion within wide limits is called racing or 
hunting. The result is that, in practice, truly isochronous gover- 
nors are not suitable to use, but that designers of governors gener- 
ally try to attain as great a degree of isochronism as they believe 
to be feasible, without introducing racing or hunting. 

In most designs of isochronous governors the frictional resistance 
at the collar and elsewhere is disregarded, and in describing the 
forms of truly isochronous governors the value of P is most fre- 
quently assumed to be either zero or a constant. To design a 
governor which should be absolutely isochronous while assuming 
P to vary, would at least lead to great complexity, though many 
approximately isochronous governors are designed and used, 
where P is, and is assumed to be, a variable, especially in cases 
where the force exerted by a spring is employed as a part of the 
value of P. 

Parabolic Pendulum Governor. 

If we take the case of a governor like that shown in Fig. 101, 
when a = c = 0, so proportioned that I = m, and hence that i' = 
i, and let the links be attached to the balls by forks, so that, if 
B = weight of each ball, and b = CA = distance from C, the 
point of suspension, to the center of the ball, we shall have I = 
m = b. 

If, now, we neglect all consideration of the centrifugal force of, 
and of the weight of, the links, and also of the rods, then the mo- 
ment equation reduces to 

a 2 

— Bb 2 cos i sin i — (B -f- P) b sin i, . . . . (1) 

9 

and this may be written 

CE = 2bcosi = mi + ^) (2) 



a 



B 



If, now, we denote by h the vertical distance of the centers of the 
balls below C, which may be called the height of the governor, 
we shall have 

h = b cos i, and h = ~ ( 1 + -^ j . . . . (3) 

This equation (3) does not contain b, but, moreover, it only holds 
even approximately, when %' — i. It gives the relation between h 
and a, and shows that h is inversely proportional to the square 
of a. 

Another conceivable, though not a practical, arrangement is 
shown in Fig. 106, where the collar is hung by vertical rods from 
a plate DD' which rests on top of the balls, and moves up and 
down with them. In this case the moment equation is 

a 2 ( P\ 

— Bb 2 cos i sin i = ( B + — J sin i , . . . . (4) 



GOVERNORS 



163 



and this may be written 
h 



CR = bcosi=^(l + 



2 bp 



(5) 



this being of the same form as equation (3), the difference being 

p 

that 7j takes the place of P. 

If, in either (3) or (5), we neglect 
P, i.e., write P = 0, we should obtain 



or 



(6) 




Fig. 106. 



which is the height of a simple con- 
ical pendulum running at a speed of 
a radians per second. 

In all three cases, the centers of 
the balls move on an arc of a circle 
whose center is C, and the center 
lines of the rods are normals to the 
path of motion of the centers of the 
balls. 

We could obtain the same re- 
sults theoretically if, as shown in 
the figure, we were to omit the rods and let the balls move out- 
wards and inwards on a circular guide BOB' rigidly fixed to the 
spindle at 0, and revolving with it. This arrangement would not 

be good from a mechanical point 
of view, and is only given to illus- 
trate the principle. In practice some 
other mechanism would have to be 
employed to accomplish the same, 
or nearly the same, result. When 
the guide is circular the ordinary 
arrangement of rods CA and CA', 
suspended at C, accomplishes it per- 
fectly. In this case, when the guide 
is circular the value of h, or CR, 
would be different for every different 
position of the balls, and hence for 
every different value of a. 

Moreover, observe that h = 

CR is the subnormal of the circle 

EE' , i.e., it is the distance, measured along the center line of the 

spindle, between the foot R of the ordinate AR and the point C, 

where the normal AC intersects the center line of the spindle. 

If, now, for the circular guide BB r , we substitute one of such 
form that the curve along which the center of the balls moves 
is a parabolic arc, with its axis vertical and coinciding with the 
center line of the spindle, its vertex being at the lowest point, the 




Fig. 107. 



E 
— Y 



164 DYNAMICS OF MACHINERY 

governor will be isochronous for the cases to which equations (5) 
and (6) respectively apply. To prove that the governor will then 
be isochronous in the cases to which equations (5) and (6) apply, 
proceed as follows: 

Let OX be the center line of the spindle, and E the center of 
one of the balls. The center of the ball is constrained to move 
along the parabolic arc OE, having its vertex 
at and its axis vertical. Draw EF normal 
to the parabola at E and cutting the axis at F. 
The vertical height of F above E or h = FN 
is .the height of the governor, and, in the case 
of the parabola, this is constant; for we have 

h=FN=NEiwiFEN=y^. . (7) 

Moreover, the equation of the parabola in this 
position will be of the form 

y 2 = 4 ax. 
Fig. 108. Then 

dy 
h = y -p = 2 a = twice the focal distance. ... (8) 

Hence this governor is isochronous; for, wherever the ball is, the 
height is the same, and since h is constant a is also constant, and 
the forces acting on each ball will be in equilibrium only when the 
speed of the spindle is a, while for any other speed of spindle the 
governor balls will be constantly moving outwards or inwards. 

To insure the balls moving on a parabolic arc, they may either 
be guided by a parabolic guide, on which they are constrained to 
move, or they may be hung by a flexible spring, from a cheek HL 
which has the form of the evolute of the parabola. 

To plot the parabola for a given speed a, proceed as follows: 
(a) Determine the value of h from the equation (3), (5), or (6), 
according to which case is considered; then (6) write out the equa- 
tion of the parabola in terms of h, and since h = 2 a the equation 
y 2 = 4 ax becomes 

(9) 








y 2 = 2 hx, 


or 

the] 


i plot the curve. 


X 2h' 



(10) 



Cross-armed Governors. 



One method of constructing an approximately parabolic governor 
is to build a cross-armed governor proportioned as follows, viz.: 

The speed a in radians per second for which the governor is to 
be designed being given, compute the corresponding value of h 
by means of equation (3), (5), or (6), according to the kind of 



GOVERNORS 



165 




Fig. 109. 



governor required. Then hav- 
ing found h, proceed to draw 
the parabola OE' whose equa- 
tion is y 2 = 2 hx. 

Next select the portions A Ax 
and A' Ax which will be in 
most common use. 

Draw normals at A and Ax 
on the left, and at A' and Ax 
on the right. Then will H' ', 
the point of intersection of the 
first two, be the point on the 
arm HH' from which is to be 
hung the rod H'A, while H, 
the point of intersection of the 
normals at A' and at Ax, will 
be the point on HH' from which 
to suspend the rod HA' . 

The mechanical details of 
construction of the rods so that 
they can be used in these positions will not be explained here. Such 

a governor will be approximately 
parabolic. 

Underhung Pendulum Governors. 

While there are many forms of 
pendulum governors, differing from 
each other in details, the same prin- 
ciple applies to all, viz., that the 
forces acting on one ball and rod 
combined must be in equilibrium 
when the governor is running at a 
constant speed, when the load on 
the engine, and the steam pressure, 
are constant and when the load is 
that corresponding to the speed. 

By way of illustration, a diagram- 
matic view of the form of the Proell 
governor will be given (Fig. 110). 
In this governor the rods, which 
F *S- 11°- are shown as bent rods in the figure 

instead of being attached to arms on the spindle, are attached to the 
collar on which rests a load, while the links are hung from arms 
on the spindle. Its action will be evident from the cut. 

Pendulum Governors with Springs. 

In these governors, that part of the moment of the centrifugal 
force that is not balanced by the moment due to the resistance at 




166 



DYNAMICS OF MACHINERY 



the collar is usually balanced by springs. Sometimes these 
springs act directly on the collar, sometimes directly on the centrif- 
ugal weights, and sometimes both kinds of springs are used. In 




Fig. 111. 

many cases the attempt is made to so construct these governors 
that they shall be approximately isochronous, the approach to 

isochronism being attained, not 
by making any attempt to keep 
the height of the governor, i.e., 
the vertical distance of the center 
of the ball from its point of 
suspension, constant, but by so 
adjusting the scale of the spring 
that when h varies P shall vary 
in such a manner that, notwith- 
standing the variation of h, the 
value of a shall remain nearly 
constant. How this is done will 
be explained more fully later, and 
especially in connection with the 
discussion of flywheel governors 
in all of which springs are em- 
ployed. 

By way of illustration, diagram- 
matic cuts will be given (a) of the 
Hartnett (Fig. Ill), and (6) of 
the Lombard governor (Fig. 1 12) . 
In the Hartnett governor (Fig. 
Ill), if we neglect all considera- 
tion of the centrifugal force of, and of the weight of, the equal-armed 
right-angled bell-crank levers to which the balls are attached, and 




Fig. 112. 



GOVERNORS 167 

if we let 

B = weight of each ball, 
n = HA = HD = H'A' = H'D' 
i = angle made by HA with the vertical, 
c = CD, 
r = perpendicular distance of A from the spindle, 

r = c + ri + ri sin i, 
P = pressure on spring, 
we obtain 

Centrifugal force of ball = —Br = — B(c -\- ri -\- r ± sin i); 

p 

Load at D = — • 

If we now take moments about H, the moment equation becomes 

a 2 P 

— Br (ri cos i) = — (ri cos i) + 2? (ri sin t), 

or, if we neglect the last term, inasmuch as sin i is small, we have 

a 2 P 

— Br ri cos i = — ri cos i, 



a 



-Br =^ 

g 2 



i.e., the centrifugal force of the ball is nearly equal to the load at D. 
Now if r = the value of r when HA is vertical, and if we sub- 
stitute r for r in the last equation, thus neglecting r x sin i as being 
small, we have approximately 



2 9 P 



2Br 

If we now let S = scale of spring, and employ one of such scale 

p 

that P = Sr , and hence S = — , we shall have approximately 

2 9 S 

a = 2W 

and as this is a constant a will be nearly constant, and hence the 
governor will be nearly isochronous. 

In the Lombard governor (Fig. 112), flat springs serve for both 
the loading and the mechanism of the governor proper. The 
vertical coil spring A on one side serves to vary the loading, and 
hence the normal speed of the spindle. A dashpot is generally 
used to dampen vibrations. 

When, as is frequently the case, it is used as a water-wheel 
governor, it controls an auxiliary motor which either operates the 



168 DYNAMICS OF MACHINERY 

gate or else which controls another larger auxiliary motor which 
operates the gate. 

Astaticity. 

Professor Dwelshauvers-Dery calls attention to the fact that, 
since the conditions of isochronism are determined by neglecting 
any considerations of the frictional resistance of the governor, it 
follows that these resistances render it impossible to secure real 
isochronism in practice. He therefore recommends that governors 
be made what he calls astatic, and his definition of astaticity may 
be explained as follows, viz. : 

Referring to the discussion on pages 150 and 151, let 

n\ = greatest speed with collar in highest position. 
n 2 = least speed with collar in highest position. 
rti" = greatest speed with collar in lowest position. 
n 2 " = least speed with collar in lowest position. 

Then he defines an astatic governor as one which is so designed 
that 

n 2 = n{ . 

If it is so designed, this speed is one that it is possible for the 
governor to have, whatever the height of the collar, whereas if 
n 2 > n-i" then it is not possible for the governor to have the same 
speed for all positions of the collar. 

The definition as he expresses it is as follows: 

Governors applied to machines are called astatic when there 
exists only one single value of the speed for which the collar can 
have any position whatever in the extent of its entire travel, and 
whatever be the direction in which the resistance acts. 

Flywheel Governors. 

These governors have their axes horizontal, are usually placed 
on the main shaft of the engine, and, in a steam engine, they gen- 
erally control the admission of steam by varying the angular 
advance of the eccentric. 

Figs. 113 and 114 show, in outline, one of the forms of such 
governors. The rods AB and A'B', to which are attached the 
swinging weights W and W (W = W / ), are jointed at A and A' 
to the arms of the governor pulley. The springs DE and D'E' 
are attached to the rim of the pulley at E and E', and to the rods 
at D and D', which points may or may not be on the lines AB 
and A'B' respectively, and the tension of these springs serves 
to counterbalance the centrifugal force of the rods and swinging 
weights. The links be and b'c' are connected at c and c' to the 
collar cc', which is loose upon the shaft, while to this collar is fixed 
the eccentric, which consequently turns with it. The action of 
the governor is as follows: 

If the speed of the shaft increases, the swinging weights with 



GOVERNORS 



169 



their rods are thrown outwards by centrifugal force, thus the collar 
is turned, the angular advance of the eccentric is changed, and the 




Fig. 113. 

cut-off is shortened, and vice versa, when the speed of the shaft 
decreases. 

Let OA = r, OC = r h AD = r 2 , Ab = m, be = I. 

W\ = weight of rod AB and its attached swinging weight. 

Wi = weight of link be. 

Wb = weight of end b of link. 

Let the point G be the center of gravity of the upper rod and 
swinging weight combined. 

Let Xi = projection of A G on AB. 

2/i = projection of A G on line through A perpendicular to AB. 
i = angle OAB. 
i f = angle OAb. 
i" = angle beA . 
S= tension of the spring. 



i'" = angle EDA. 
d = angle cOA' . 
a = angular velocity of shaft in radians per second. 



7T 



N 



30 



iV = number of revolutions of shaft per minute. .'. a = 
P = pull in link along cb. 

The relation between %' and i" may be found as follows, viz.: 
1° From triangle AOc we have 

n 2 = r 2 +Ac?-2r (Ac cos CAO). 



170 



DYNAMICS OF MACHINERY 



and 



2° From triangle A be we have 

Ac 2 = m 2 + l 2 + 2lm cos (£' + i"), 

Ac cos CAO = m cos i' + I cos i" . 

Hence by substitution 

r i 2 = r 2 -\- m 2 -\- 1 2 + 2 Im cos (i'+i") — 2r cos i' (m cosi'+lcosi"). (1) 

By making use of this equation we can, in any actual case, plot a 
curve having values of i' for abscissae and values of i" for ordinates. 




Fig. 114. 

Now take moments of the forces acting on the upper rod and 
its swinging weight combined, about an axis through A at right 
angles to the plane of the pulley, and let 

Mi = moment of centrifugal force of W\. 

M 2 = moment of force exerted on AB, in consequence of cen- 
trifugal force of link be. 

Mz = moment of tension in spring DE. 

M 4 = moment of force exerted on AB by the pull along the 
link be. 

Then, in order that the forces acting on A B may be in equilibrium, 
we have 

Mi + Mt = Ms + M i} (2) 

and this is the moment equation. It is necessary, however, to 
work out the values of the several moments, and substitute them 



GOVERNORS 171 

in (2). Before undertaking this, for the general case, a special 
case will be worked out in an approximate manner. 

CASE I. 

(a) Disregard the centrifugal force of the link. Hence write 
M 2 = 0. The moment equation (2) then reduces to 

M 1 = M 3 + M if (3) 

which means that the moment of the centrifugal force must 
balance the sum of the two moments (1°) that of the spring ten- 
sion and (2°) that of the resistance P. 

(b) Let the point b coincide with B. Hence we have i' = i. 

(c) Assume that the governor is so constructed that in ordinary 
use the angle ABc does not differ greatly from a right angle, and 

as an approximation write ABc = - • .*. i" = - — i. 

(d) Assume the spring to be so located that in ordinary use 
the angle EDA does not differ greatly from a right angle, and 

as an approximation write EDA = - - Hence %'" = ~ • We then 

2i & 

have 

M 2 = 0, M 3 = Sr 2 , and M 4 = Pm. 

To deduce M h proceed as follows, viz. : 

Let AP = perpendicular from A on OG. 

GQ = perpendicular from G on OA = projection of AG on 
a line at right angles to OA = xi sin i + y\ cos i. 
Then we have 

a 2 

— W\ (OG) = centrifugal force of Wi, 

g 

and this acts along the line OG. Hence 

Mi = - Wt (OG) (AP) = — Wi (twice area of triangle OAG) 



9 g 

Wi(OA)(GQ), 



a 2 



g 

But OA = r, and GQ = Xi sin i -\-yi cos i. 

a 2 
Hence M i = — Wir (xi sin i + yi cos i) (4) 

This value for Mi is correct for the general, as well as for this 
special, case; the values of M 2 , M z , and M 4 given above, however, 
apply only to this special case. 

The moment equation for this case, therefore, becomes 

a 2 

— Wir (xi sin i -\-yi cos i) = Sr 2 + Pm. ... (5) 

g 



172 DYNAMICS OF MACHINERY 

Examples. 
In all the following examples let 

Wi = 17.74 pounds, Xi = 7.73", y x = 0.51", m = 16", 
r 2==AD = 4#87 // > r = OA = 12.8", Also, a = ^ • 

The moment equation therefore becomes, when this value is sub- 
stituted for a, 

ir 2 N 2 r 
(900) (386) Wl ^ Xl Sm * "^ Vl C0S ^ = ^ r2 ~*~ jPm ' ' ' ^ 
and if we substitute the data given above, it reduces to 

N 2 (0.04982 sin i + 0.00329 cos i) = 4.87 S + 16 P. . (7) 

Equation (7) may therefore be used in solving the following 
examples : 

Example 1. —Given i = 28° 30', N = 117, and S = 10 pounds, 
find P. Result: P = 19.77 pounds. 

Example 2. — Given i = 28° 30', N = 117, and S = 20, find P. 
Result: P = 16.72 pounds. 

Example 3. — Given N = 150, i = 28.30, and £ = 40, find P. 
Result: P = 25.32 pounds. 

GENERAL CASE. 

In the general case, as has been already shown, we must have 

Mi + M 2 = M 3 + M 4 . 
1° To deduce Mi, proceed as follows: 

Let AP = perpendicular from A on OG. 

GQ = perpendicular from G on OA = projection of AG, 
on a line at right angles to OA = Xi sin i -f- 2/i cos i. 
Hence 

a 2 

— TFi (OG) — centrifugal force of Wi, and this acts along OG. 

Therefore 

Mi=-Wi (OG) (AP) = -Wi vtwice area of triangle OAG) 
y «/ 

= jWdOA)(GQ)- 

But OA = r, and GQ = #i sin i + 2/1 cos i. 

a 2 
Hence Mi = — Wtf (xi sin i + y x cos i). 

2° To deduce M 2 , proceed as follows, viz. : 

Instead of determining the centrifugal force of the entire link, 
which of course acts along the line joining with the center of 



GOVERNORS 173 

gravity of the link, and resolving this into two components, one 
along Ob, and the other along Oc, we shall arrive at the same 
result by a shorter method, if we substitute for the link two con- 
centrated weights, one at b, and one at c, the center of gravity of 
the combination being the same as that of the link. The magni- 
tude of the weight at b will be Wb, i.e., the weight of the link at b, 
and that of the weight at c, will be W c , i.e., the weight of the link 
at c. The centrifugal force of W c acts along Oc and only causes 
stress in the collar, being opposed by the corresponding force 
arising from the link on the other side of the shaft. Hence the 
centrifugal force of the weight Wb at b, which acts along Ob, is 
the only force due to the centrifugal force of the link that has any 
effect on AB. 

Let APi = perpendicular from A on Ob, and 

bQi = perpendicular from b on OA = projection of A b, on 
line at right angles to OA = m sin i' . 
Hence we have 

a 2 

— Wb (Ob) = centrifugal force of Wb, and it acts along Ob. 

Therefore 

M 2 = - W h (Ob) (APi) =-W b (twice area of triangle OAb) 

if a 

= -W b (OA)(bQ 1 ). 

But OA = r, and bQ = m sin i' '. 

a 2 
Hence M 2 = — Wb rm sin i'. 

9 
3° To deduce the value of M 3 . 

This is simply the product of the spring tension S by the per- 
pendicular from A upon its line of action. Hence 

M z = Sr 2 sin %'" . 

4° To deduce the value of Af 4 , proceed as follows: 
Let P = pull in the link. Then we have from the figure that 
the perpendicular from A on be is m sin (i r + i"). Hence 

M 4 =Pmsin(i' + i"). 

Hence, substituting these quantities in equation (2), we obtain 
for the moment equation of a flywheel governor of this type 

— \W\ (xi sin i + wi cos i) + Wb m sin i' \ 
Q I ) 

= Sr 2 sin %'" + Pm sin (i' + i") (8) 

Moment Equation of Flywheel Governor when the Axis of x is Taken 
as the Line Joining A with the Center of Gravity of Wi. 

Whenever Wi does not consist of a rod, and added weight, but 
is a body of some other shape, whether symmetrical or not, it will 



174 



DYNAMICS OF MACHINERY 



generally be found more convenient to take the axis of x as the 
line joining A, the pin, with the center of gravity of W\. 
In this case, y\ = 0, and hence equation (8) reduces to 



a z r 



— { WiXi sin i + Wb m sin i' \ 
9 



Sr 2 sin %'" + Pm sin (%' + i"). (9) 



Moment Equation of Flywheel Governor Having but One Swinging 

Weight. 

This case, which includes most so-called inertia governors, may 

be subdivided into two 
cases, as follows, viz. : 

1° That in which no 
eccentric is used, but 
where the eccentric rod 
is attached to a wrist 
pin rigidly fixed to the 
swinging weight, as 
shown in Fig. 115. 

2° That in which the 
eccentric is firmly fixed 
to and forms a part of 
the swinging weight, as 
shown in Fig. 116. 

In both cases the term 
Wb m sin i' of equation 
(9) vanishes. Hence for 
the first case, equation 
(9) reduces to the form 




Fig. 



115. 



a z r 

g 



WiXi sin i = Sr 2 sin i'" + Pm sin (i' + i"), 



(10) 



where P = force exerted along the eccentric rod. 

In the second case, however, a part of the moment of the resist- 
ance is made up of the moment of the friction of the eccentric 
strap on the eccentric, and hence it cannot be expressed in the 
form Pm sin (i f + i"). With this in view the equation (9) can 
only be reduced to the form 



otr 



— TTiXi sin i = Sr 2 sin %'" + M 4 . 

g 



(11) 



A very general form, which can be made to include all the cases 
when the axis of x is taken as the line joining the pin A with the 
•center of gravity of W h is the following, viz. : 



g 



\ WiXi sin i + W b m sin i' j = Sr 2 sin i'" + M 4 . 



(12) 



GOVERNORS 



175 



Approximate Forms of the Moment Equation. 

If, as an approximation, we neglect all consideration of the 
centrifugal force of the links and hence write Wb m sin i = 0, 
consider i'" as a right angle, and hence write sin %'" — 1, and, if 




Fig. 116. 

using equation (9), we consider the link at right angles to the axis 
of x, and hence write 



2 



i' + i 



,*// 



7T 



then equation (9) becomes 



a z r 

g 



WiXi sin i = Sr 2 + Pm, 



(13) 



which will be most convenient for a case with two swinging weights. 
On the other hand, if we make the same substitutions in equation 
(12), we obtain 



a z r 
9 



WiXi sin i = Sr 2 + M 4 , 



(14) 



which will be most convenient for a case with only one swinging 
weight. 

Moreover, equation (14) includes equation (13) if we observe 
that in that case M 4 = Pm. 

Value of the Angle 6. 

In designing a governor similar to that shown in Fig. 114, or 
in endeavoring to predict the behavior of such a governor already 
constructed, it is necessary to study the relation between (1°) the 



176 DYNAMICS OF MACHINERY 

position of the collar cc', (2°) the speed of the governor shaft 
and (3°) the value of P. The position of the collar cc' is determined 
when we know the angle i, but it is often desirable to use the 
angle = cOA' ', and therefore to express 8 in terms of i' and i", 
so that when these are known it can be computed. To do this, 
consider the triangle cOe. From it we have 





7 sin % 


sin 


ec I — be "sin i" I sin i" — m sin %' 


sin i" 


Oc ri r x Ti sin i" 

. _ I sin i" — m sin i' 



(15) 

Calculations of a similar nature will serve to determine the relation 
between the angular advance of the eccentric and the angle i in 
other forms of flywheel governors. 

General Discussion. 

In discussing the action of any given flywheel governor of the 
kind shown in Fig. 114, we can start with the three equations 

n 2 = r 2 + ra 2 + I 2 + 2 Im cos (i' -f i") — 2 r (ra cos i' + 1 cos i") (16) 
— ] W\ (xi sin i + 2/1 cos i) + "FT^m sin i > 

= Sr 2 sin i ,,, + Pm sin (t v + i"), . . . (17) 

. _ I sin i" — m sin i' _- N 

sin = (18) 

Equation (1) gives the relation between i" and 2 V , so that for any 
given value of i' the corresponding value of i" can be determined. 
This in most cases will have to be done by successive trials. 
Then one way to proceed would be to plot a curve, with the values 
of i' as abscissae, and the values of i" as ordinates ; and then an 
empirical equation derived from the curve can be formed which 
will give i" in terms of i' . 

The value of i' can easily be expressed in terms of i, since i' = 
i — bAB = i — a constant. 

Hence we may say that in equation (2) we have the following 
four variables, viz., i, a, S, and P. Of these four, however, S 
can be expressed in terms of i as soon as we know the scale of the 
spring and its initial tension, S , corresponding to some value 
io of i; for if e = elongation of spring corresponding to the change 
of angle from i to i, and if a is the scale of the spring, we have 

S = So + ae. 

Hence we may say that we have in equation (2) only three variables, 
viz., i, a, and P. Now, if any two of these are known the third 



GOVERNORS 177 

can be determined, and, on the other hand, if only one is known 

we can obtain an equation giving the relation between the other 

two, or this relation can be represented by a plane curve. As to 0: 

that is known as soon as i is known, and so also N, the number of 

wN 
revolutions of the governor shaft per minute, since a = — . Hence 

we may have any one of the following three cases : 

1° Given i or 0, to determine the relation between P and a, or, 

which amounts to the same, between P and N. 
2° Given a or N, to determine the relation between P and i, 

or, which amounts to the same, between P and 0. 
3° Given P, to determine the relation between i and a, or, 

which amounts to the same, between and a, or between 

and N, or between i and N. 

Scale of the Spring. 

Besides the above, we may study the effect of varying the scale 
of and the length of the spring, this study being very important, 
inasmuch as the choice of a suitable spring affects very much the 
action of the governor. 

In the case of the ordinary flywheel governor, when the swinging 
weights are farthest out from the center of the shaft, and hence 
when the spring has its greatest length, the speed of the governor, 
and hence that of the engine, has its greatest value, while the cut-off 
is shortest. On the other hand, when the swinging weights are 
nearest the center of the shaft, and hence the spring has its least 
length, the speed of the governor, and hence that of [the engine, 
has its least value, while the cut-off is longest. For any one cut-off, 
and hence for any one value of i, and therefore for any one length 
of spring, there is a definite speed, which can be determined from 
the moment equation; and at this speed the engine must run, 
in order that equilibrium may exist between the forces acting on 
the swinging weight. 

Moreover, the speed corresponding to any one cut-off, and 
hence to any one value of the angle i, is different from the speed 
corresponding to a different cut-off, and hence to a different value 
of the angle i. 

Moreover, if iV max . = greatest speed, i.e., speed corresponding 

to shortest cut-off, and 

if A^ m in. = least speed, i.e., speed corresponding to 
longest cut-off, 

then it is desirable to so construct the governor that iV max . — Nwm. 
shall be as small as possible, provided we do not make it so small 
that racing or hunting is introduced. 

The magnitude of iV max . — A^mm. can be controlled by a suit- 
able choice of springs. If these were so chosen, however, that 



178 DYNAMICS OF MACHINERY 

Nmax. — N m \ n . were reduced to zero, the governor would be isochro- 
nous, but in that case it would be too sensitive, and would race. 

Before discussing, however, the approximate limits of variation 
of speed, we will first ascertain the conditions required to produce 
approximate isochronism. 

Isochronism. 

To compute the scale of the spring that would produce isochro- 
nism, provided the conditions justify us in making use of the approx- 
imate moment equation (14) and in assuming ikf 4 to be constant 
for all values of i, proceed as follows, viz. : 

Equation (14) may be written 

a 2 

— WiX! r sin i = Sr 2 + ikf 4 (17) 

9 

If the governor is to be isochronous, then a must remain constant 
for any change di in the angle i and the consequent change dS 
in the tension on the spring. Hence by differentiating (17) we 
obtain 

a 2 

— WiXir cos i di = r 2 dS (18) 

9 

Let a be the scale of the spring; then since its elongation when the 
angle i increases to i + di is r 2 di, we have 

• dS = ar 2 di (19) 

Substituting the value of dS in equation (18), we obtain 

a 2 

— WiXi (r cos i) di = ar 2 2 di (20) 

9 

And if we divide out by di, and for r cos i substitute x , where xo 
denotes the distance from A to the foot of the perpendicular 
dropped from on the axis of x, we obtain 

a 2 W1X1X0 , . 

a = =— , (21) 

g r 2 2 

and this is the scale of spring required to produce isochronism. 
Hence in the actual governor a larger scale of spring should be 
used; nevertheless, the nearer the scale approaches the value given 
in equation (21) the more sensitive will the governor be. 

Approximate Limits of Variation of Speed. 

The following approximate method of studying the limits of 
variation of speed of a flywheel governor not only adopts all the 
approximations and assumptions that result in equation (14), 
but also neglects ikf 4 , the moment of the resistance in that equa- 
tion. Hence the method is very inexact. 

The moment equation used is, therefore, 

- TTiXi (r sin = Sr 2 , (22) 

*7 



GOVERNORS 179 

or if we let p = r sin i = the length of the perpendicular from 
on the axis of x, and let x = r cos i = distance from A to the foot 
of the perpendicular from on the axis of x h we have for the 
moment equation 

-TTi^p = Sr 2 (23) 

Let S be the tension of the spring corresponding to the mean 
angular velocity a. 

Suppose that we require that the extreme variation of speed 

shall not be more than — th of the mean speed, and that we are to 

m 

determine the total difference between the extreme distances from 

the center of the shaft, of the foot of the perpendicular P from 

on the axis of x. The greatest speed is a(l+- — ), and to this 

corresponds the greatest distance, which will be denoted by p + K\) 
also the greatest tension in the spring, which will be denoted by 

$ + Ai$. The least speed is f 1 — - — y and to this corresponds 

the least distance, which will be denoted by p — K 2 ; also the least 
tension of the spring, which will be denoted by S — A 2 S. Then, 
if we denote by K the total difference of distance, we shall have 

K=K 1 +K 2 (24) 

Moreover, from (23) we derive the two following equations, viz.: 

ry WlXl „ 

o = — ocp. 

qr 2 

S + AtS = ^ « 2 (l + J-Y(p + K,), 
gr 2 \ 2m) " 

gr 2 (\ 2 m) ^) 

The elongation of the spring = Ki— • Hence if the scale of the 

Xo 

spring is a 7 we have 

A 1 S = aK 1 - (26) 

Xq 

Hence, equating (25) and (26) and solving for K h we obtain 



WiXiXqol 2 \ 2m, 



180 



DYNAMICS OF MACHINERY 



By a similar process we should obtain 

1^ 1 

m 



K 2 

Hence we have 

K = K 1 +K 2 



= P 



4 m 2 



agr 2 l 



WiXiXqO 2 



V 2m) 



(28) 



1 

m 


+4 1 . 

4 m 2 




agr 2 2 

WiXiX a 2 


-d i 


i y 


V ' 


2 m) 



+ 



1 

m 


1 

4 m 2 


agr 2 2 
WiXiX a 2 


\ 2 m) 



(29) 
If, in the denominators, we omit -= — as being small compared with 1, 
we obtain 



2 

m 



agr 2 2 



- 1 



K = P 

TTi^iXoa 2 
This equation shows that we must have 

WiXiXq 



(30) 



agr 2 z 



TFi^i^oa 2 



1 > 0, and 



a > 



gr 2 l 



a 2 , otherwise K would be negative and the governor 



would not perform its functions properly. 

When a spring is used whose scale is a = — ^—^ a 2 , then the 

gr 2 2 

denominator of (30) becomes zero, and since K must be finite we 

would have — = 0, and consequently the governor is isochronous 

if this particular spring is employed. Moreover, this is the same 
condition as that given in equation (21) for isochronism. In a 
practical governor the scale of the spring must be greater. 



Inertia Governors. 

Confining ourselves to flywheel governors, which have only 
one swinging weight, the moment equation is 



a 



— Wirxi sin i = Sr 2 sin %'" + M4, 

g 



(i) 



which has already been deduced in equation (11), page 174. For 
convenience, however, a new figure will be drawn applicable to 
this class of governors, and the following lettering will be adopted, 
viz.: 



GOVERNORS 



181 



will denote the center of the shaft. 

P will denote the center of the pivot about which the swinging 

weight turns. 
G will denote the center of gravity of the swinging weight. 
E will denote the center of the wrist pin, or the center of the 

eccentric, according 

to which is employed. 
i = GPO. 

We shall then have 

r = OP. 
xi = PG. 

r 2 sin %'" = perpendicular 
from P on the spring; 
also, 
£ = x\ sin i = perpen- 
dicular distance of G 
from OP. 

Substituting £ for xi sin i 
in equation (1), it becomes 
a 2 




L - J 



W 1 rZ = Sr 2 8mi'"+M A .(2) 

Moreover, the quantities 

denoted by the separate Fig. la- 

terals of equation (2) are as follows, viz. : 

— W\r% = moment, about P, of centrifugal force of swinging 

«7 

weight. 
Sr 2 sin %'" = moment, about P, of spring tension. 
M 4 = moment, about P, of resistance. 

Graphical Representation of the Moment Equation. 

A convenient way of studying the action of the governor under 
constant load, when the moment equation is known, is as follows: 

First determine, from the 



o^- 



2*C-— 



M 1 1 

Mil 



(Xc 



I i i I 



I "I 



Values ot ?■ 



M 4 



dimensions of the governor 
and valve gear, the values 
of £ corresponding to certain 
definite cut-offs, as 0.1, 0.2, 
0.3, 0.4, 0.5, 0.6, 0.7, etc.; 
then plot a curve with these 
values of £ as abscissae, and 
with ordinates equal to the 
corresponding moment of 

mean resistance. This curve is marked M 4 in Fig. 118. 

Next determine by graphical construction, or otherwise, the 

length of the spring, and the length of its moment arm about P, 



.7 .6 .5.1.3.2.1 Cut off 



Fig. 118. 



182 DYNAMICS OF MACHINERY 

corresponding to each of the cut-offs, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6,, 
0.7, etc., and from these data, and the results obtained by cali- 
brating the spring, determine the moments of the spring tensions,, 
at the same cut-offs. 

Then add the ordinates representing these moments of the 
spring tensions to those of the curve marked M 4 , Fig. 118, re- 
spectively, and thus obtain a new curve (M f in Fig. 118). The 
ordinates of this new curve will represent, to the same scale, the 
moments of the centrifugal force required to balance the spring 
tension, and the mean resistance combined; hence these ordinates 

will represent the values of the quantity a 2 — r£. Moreover, 

since W and r are constants, we can readily compute the value of a 
corresponding to any given value of £, and hence to the corre- 
sponding cut-off. Then values of a can then be plotted (using a 
convenient scale) as ordinates, and thus the speed curve can be 
obtained which is marked a in Fig. 118. 

It is necessary, in order that the governor may be stable, that, 
an increase of the displacement of the swinging weight, i.e., an 
increase of £, should correspond to an increase of speed and a dimi- 
nution of cut-off. 

If through a point on the centrifugal force curve (as, for instance,, 
that corresponding to 0.5 cut-off) a " ray " be drawn from 0, the 
point for which £ = 0, and which corresponds to the shaft center, 
or, rather to the line OP, Fig. 117, then will any ordinate of this 
ray (as, for instance, that corresponding to the value of £ for 0.3 
cut-off) represent the moment of the centrifugal force that the 
swinging weight would have, were this value of £ the abscissa of 
the point, and were the speed the same as that corresponding to 
the point on the centrifugal force curve through which the ray is, 
drawn (in this case 0.5 cut-off). Moreover, the difference between 
the ordinate of the curve at say 0.2 cut-off, and that of the ray 
for the same abscissa, will represent the unbalanced moment of 
centrifugal force, acting on the swinging weight, at the instant 
when, with the engine having run steadily thus far at 0.5 cut- 
off, the load is suddenly changed to one corresponding to 0.2 
cut-off. 

It is necessary for stability that no tangent to the centrifugal 
force curve shall pass through 0, and that all tangents should 
cut the axis of ordinates on the same side of 0. 

General Remarks on Governors with only One Swinging Weight. 

When the governor shaft is revolving at a constant speed, and 
the load is constant, and is the one corrresponding to that speed, 
the only forces (if we neglect the effect of gravity) tending to make 
the swinging weight change its position relative to the governor 
wheel, and hence to perform their part in the regulation of the. 



GOVERNORS 183 

engine, are (1°) centrifugal force, (2°) spring tension, and (3°) resist- 
ance to the motion of the valve gear; and under the circumstance 
stated above these forces must be in equilibrium, so that no rela- 
tive motion of the swinging weight may occur. The analytical 
expression of this condition is the moment equation. 

A study of the action of the governor by means of the moment 
equation is often called the statical treatment of the governor, 
inasmuch as it deals with its action under a constant load, and at 
a constant speed; and, when the load is changed, it only deals 
with the conditions that will result in a constant speed under the 
new load. When, however, as under the influence of a change of 
load or a change of steam pressure, the speed of the governor 
shaft changes, and hence that of the governor wheel, the tendency 
of the swinging weight to continue at the same speed in consequence 
of its inertia comes into play, and as a consequence accelerating 
forces are developed, which will aid or injure the regulation, accord- 
ing as, in the design and construction of the governor, they have 
been properly or improperly arranged. 

In the preceding discussion of governors, i.e., that by means 
of the moment equation only, no attempt has been made to 
take these forces into account, as they only act during the time 
while the governor is passing from one to another condition of 
equilibrium. 

In all governors, whether pendulum or flywheel, inertia forces 
come into play when either the load or the steam pressure changes, 
but the class in which it is most necessary to consider them is in 
that of flywheel governors, which are purposely designed in such 
a manner that as much benefit as possible may be derived in the 
regulation from these inertia forces. Such governors are often 
called inertia governors. Many of them consist of a flywheel 
governor with only one swinging weight, which is purposely made 
comparatively large, while the distance OP is made correspond- 
ingly small, the swinging weight serving both to develop centrifugal 
force and inertia. 

In some, the swinging weight intended to develop centrifugal 
force is so designed that it does not have as much inertia as is 
desired, and another so-called inertia weight is connected with it 
by links or otherwise, and often turns around instead of P, for 
the purpose of adding to the inertia forces developed by the so- 
called centrifugal force weight. 

Other arrangements are sometimes made, but the discussion 
that follows will be confined to flywheel governors having only 
one swinging weight. A similar discussion, with some modifica- 
tions in the details, but based on the same principles, would serve 
in the case of other forms. Observe that, while inertia can be made 
to aid in the regulation, centrifugal force is necessary, and that 
any attempt to depend wholly on inertia and to do away with 
centrifugal force, as by making OP = 0, would result in failure to 
regulate. 



184 



DYNAMICS OF MACHINERY 



Arrangement of the Swinging Weight. 

If the inertia forces are to aid and not to injure the regulation, 
the arrangement must be such that, when the speed increases, 
they shall tend to shorten the cut-off, whereas, when the speed 
decreases, they shall tend to lengthen it. 

Hence in designing the governor, the arrangement, and the 
action of the valve gear, must be fully considered, and taken into 
account. Illustrations will be given of the proper arrangement, 
in certain cases of plain slide-valve engines, when the slide valve 
is driven by the valve rod, which in turn is driven by the eccentric 
rod, and this in turn by a wrist pin rigidly attached to the swinging 
weight. 



I 1 




mmmmBzw 



Fig. 119. 

Of course the same arrangements will apply to the cases when 
an eccentric is used instead of a wrist pin, this eccentric being 



^1 



'§* 




Fig. 120. 

rigidly fixed to and forming part of the swinging weight, as the 
eccentric center takes the place of the wrist-pin center. As far 



GOVERNORS 



185 



as the engine and valve gear are concerned, we must consider 

1° The direction of rotation of the crank; 

2° Whether the valve takes steam on the outside or the inside ; 

3° Whether a rocker is used to reverse the motion or not. 

In the light of the above, we must determine the general position, 
inrelation to the .crank, of the following points in the swinging weight : 

1° That of the point P of the swinging weight; 

2° That of the center of gravity of the swinging weight; 

3° That of the spring, and the direction of its pull ; 

4° That of the wrist pin E, or of the eccentric center. 

The first illustration applies to the case when the valve takes 
steam on the outside, and when no rocker is used. 

The same arrangement of the governor applies to the case when 
the valve takes steam on the inside, and a rocker is used to reverse 
the motion. 

The second illustration applies to the case when the valve takes 
steam on the inside, and no rocker is used. 

The same arrangement of the governor applies to the case when 
the valve takes steam on the outside, and a rocker is used. 

Oscillations under Constant Load. 

Were M 4 constant for each cut-off, the moment equation, or 
its graphical representation, would suffice for a study of the action 
of the governor under constant load and constant steam pressures. 
But as M 4 , for any one cut-off, varies periodically, oscillations 
occur which, in some cases, become of importance. 




Fig. 121. Fig. 122. 

They can be determined graphically, analytically, or by a com- 
bination of the two methods. 

Their determination, in what follows, while along the lines laid 



186 



DYNAMICS OF MACHINERY 



down by Mr. O. Schneider, substitutes analytical for considerable 
of his graphical work. 

Consider a vertical single-cylinder engine (Figs. 121 and 122), 
with an overhanging shaft governor, with one swinging weight, 
the eccentric rod being driven by a wrist pin, and driving in its 
turn one end of a rocker, to the other end of which is attached 
the valve rod. The resistances to be considered are: 

1° The throw of the reciprocating masses of valve and valve gear; 

2° The steam pressure on the end of the valve rod ; 

3° The weight of valve, valve rod, and eccentric rod; 

4° The valve resistance, including that in the stuffing box; 

5° The action of gravity on the swinging weight; 

6° The internal friction, especially that on the pivot. 

Of these, 1, 2, 3, and 4 all act at the wrist pin, but neither 5 nor 6. 

Considering any one of the first four, it consists of a force F 
applied at E, whose line of action is nearly or exactly parallel to 
the line of dead points of the engine. 

The path of E relatively to the governor wheel is a circle with its 
center at P, Fig. 117. 

If we resolve F into two components, one along the relative path 
of E, and the other at right angles to it, the second component is 
resisted by the pin P, and hence the first component is the only 
one that produces any motion of the swinging weight relatively 
to the governor wheel. 

A simpler, and a sufficiently accurate, method will be to assume 
the relative path of E to be a straight line perpendicular to OP y 



r 



+F 



I 



f 




Fig. 123. 

Fig. 117, and cutting OP at a point L midway between the points 
where OP is cut by the arc which forms the relative path and its 
chord. 

Notation. 

a = angular velocity of crank in radians per second. 
at = crank angle in radians, measured from line of dead points, 
starting at head end. 
t = time, in seconds, in which at is described. 
5 = angular advance of eccentric in radians. 



GOVERNORS 



187 



r = OP (Fig. 117) = distance, in inches, of swinging weight pivot 

P from shaft center. 
R — eccentricity in inches. 
ri = PL (Fig. 117) in inches. 

£ = OA (Fig. 117) in inches = perpendicular from G on OP. 
W = weight of swinging weight in pounds. 

g = 386 inches per second. 

W 

m = — = mass of swinging weight (units, pounds and inches). 
y 

= angle GPO (Fig. 117) in radians. 

x = distance, in inches, of valve from middle of its travel. 
F = variable resistance parallel to line of dead points in pounds. 
P = variable component of resistance, in pounds, along the rela- 
tive path of the eccentric for crank angle at. 
P m = mean value of P in pounds. 

M = variable moment of resistance for crank angle at in inch- 
pounds. 
M m = mean moment of resistance in inch-pounds. 
Mi = M — M m = variable moment of resistance from the mean, 

in inch-pounds, for crank angle at. 
Mi = mean moment of total resistance corresponding to one 
given cut-off in inch-pounds. 
6 = variable angular velocity of swinging weight in radians per 
second for crank angle at. 
d m = mean value of 6. 

&i = 6 — 6 m = variable angular velocity of swinging weight 
from the mean, in radians per second, for crank angle at. 
y) = variable angular displacement of swinging weight, in ra- 
dians, for crank angle at. 
r) m = mean angular displacement of swinging weight in radians. 
tqi = y] — r\m = variable angular displacement of swinging weight 

from the mean, in radians, for crank angle at. 
p = radius of gyration, in inches, of swinging weight, about its 
axis, through the point P. 

1 = Wp 2 = moment of inertia of swinging weight, about its 

axis, through the point P. 

Considering, first, any one of the four resistances that act at the 
wrist pin, since the direction of its line of action is always nearly, 
or exactly, parallel to the line of dead points of the engine, we 
shall have, if we denote its magnitude for crank angle at by F, and 
if P denote the corresponding component (also for crank angle at) 
along its relative path, that 

P = ^sincrf, (1) 

as will be evident from a perusal of Fig. 123. Hence we shall have 
for the moment of this resistance, about the point P, for the crank 
angle at, 

M = (F sin at) (PL) = (F sin at) r x (2) 



188 



DYNAMICS OF MACHINERY 



For the resistances that do not act at the wrist pin, viz., (5) and 
(6), the formulae (1) and (2) given above do not apply. A formula 
will, however, be deduced later, in the case of the action of gravity 
on the swinging weight, giving the value of M in terms of at. 

As to the resistance due to friction on the pivot, it needs to be 
determined experimentally. 

Determination and Graphical Representation, for Any 
One Cut-off, of 

(a) M, the moment of each resistance, and of the total resistance, 
for each crank angle at; 

(b) M m , the mean moment of each resistance, and of the total 

resistance ; 

(c) Mi = M — M m , the excess of the moment of each resistance 

at each crank angle at above the mean moment of that 
resistance, and the excess of the moment of the total 
resistance at each crank angle above the mean moment of 
the total resistance. 

To represent graphically the moments, about P, of the separate 
resistances and of the total resistance, proceed as follows, viz. : 

Along the axis of abscissae, and starting with the head-end dead 
point as origin, lay off a length OX, Fig. 124, to represent one 
revolution of the crank (2 t if in radians, 360 if in degrees). 




Fig. 124. 

Divide this length into thirty-six equal spaces, each representing 
ten degrees of crank angle. 

To obtain the curves representing the moments of the separate 
resistances, and that representing the moments of the total resist- 
ance, proceed as follows, viz. : 

In the case of each separate resistance lay off as ordinates, at 
each point of division, the value of M corresponding to that crank 
angle, and thus plot the curve representing the moments of that 



GOVERNORS 



189 



resistance. Do the same for each of the separate resistances. 
Having plotted these separate curves, proceed to plot the 2M 
curve whose ordinate at any one point of division is equal to the 
algebraic sum of the ordinates of the separate curves at the same 
point of division. The 2M curve, and also the separate curves, are 
shown in Fig. 124, where 

aa! is that due to the throw of the reciprocating parts of valve 
and valve gear; 

cc r is that due to the weight of valve, valve rod, and eccentric 
rod; 

dd r is that due to the friction of the valve and valve rod; 

ee r is that due to the action of gravity on the swinging weight; 

//' is the 2M curve. 
The curve due to the steam pressure on the end of the valve rod is 
absent because, in this engine, the valve takes steam on the inside. 

6 Scale 




^M curve 

6 curve 

1? curve 



Fig. 125. 

If we wish to determine the mean moment of resistance M m for 
any one of these curves, we can find it in either one of the follow- 
ing ways, viz.: (a) graphically, by determining its resultant area 
(areas above OX being positive, and below negative) by the use 



190 DYNAMICS OF MACHINERY 

of a planimeter, or otherwise, and then the quotient obtained by 
dividing this resultant area by OX will be the value of M m desired; 
or (b) we may find it analytically by means of formulae that will be 
given later for each separate resistance; and, (c) for the 2M curve, 
the mean moment M m may be found by taking the algebraic sum of 
the mean moments of the separate resistances. Having thus found 
for each separate curve, as well as for the SM curve, the value of M 
at any crank angle at, and also the mean moment M m , we readily 
obtain, by subtraction, the value of M\ = M — M m for any crank 
angle at. 

In Fig. 124, Og is laid off to scale to represent the mean moment 
M m for the 2M curve, and the line gg' is drawn parallel to OX. 

In Fig. 125 the lines OX and gg' and the SM curve are copied 
from Fig. 124, but the other curves of Fig. 124 are not drawn. 

Angular Velocity of the Swinging Weight. 

In order to obtain the angular velocity B of the swinging weight 
(for crank angle at) due to any one resistance, we proceed as follows, 
viz.: 

(a) The moment causing angular acceleration is M\ = M — M m . 

(b) The angular velocity is, by definition, 0. 

d0 

(c) The angular acceleration is consequently -r- • 

Hence, from the principles of mechanics, we have 

Mi=~; .\ 0=4 ['Midi, 

g at I Jo 

Hence, substituting for M i its value in terms of at, and integrat- 
ing between t and 0, we obtain the value of B in terms of at. 

The mean value of 0, i.e., B m , is to be obtained from the formula 



Bm = 



X2tt 
dd (at) 



TV 



as is evident from the figure. 

Observe that, in plotting these values of 6 and of B m as ordinates, 
the axis of abscissae is not OX, but the parallel line whose ordinate 
is M m . 

Having determined 6 for each of the separate resistances, their 
algebraic sum will be the value of B corresponding to the 2M 
curve. In plotting the B curve in Fig. 125, therefore, these values 
of B are laid off as ordinates from the line gg' taken as axis of abscissae. 
Moreover, the line gh which represents the value of B m correspond- 
ing to the 2M curve is found by taking the algebraic sum of the 
values of B m for the separate resistances. 



GOVERNORS 191 

Angular Displacement of Swinging Weight. 

In order to obtain the angular displacement of the swinging 
weight (for crank angle at) due to any one resistance, we proceed 
as follows: 

(a) The angular velocity above the mean is 0i = — 6 m . 

(b) The angular displacement is, by definition, yj. 

d~f\ 

(c) The angular velocity is consequently -tt • 

Hence, substituting for 0i its value in terms of at, and integrating 
between t and 0, we obtain the value of yj in terms of at. 

The mean value of yj, i.e., yj to , is to be obtained from the 
formula 

' rid (at) 

f]m = 7> t 

Ztt 

as is evident from the figure. 

Observe that, in plotting these values of yj and r\ m as ordinates, 
the axis of abscissae is neither OX nor the parallel line where the 
ordinate is M m , but the parallel line whose ordinate is 6 m . Having 
determined yj for each of the separate resistances, their algebraic 
sum will be the value of yj corresponding to the 2M curve. In 
plotting the yj curve in Fig. 125, therefore, these values of yj are 
laid off as ordinates from the line hh' taken as axis of abscissae. 
Moreover, the line hK which represents the value of yj to correspond- 
ing to the Silf curve is found by taking the algebraic sum of the 
values of yj to for the separate resistances. (In the figure, hK 
happens to be numerically negative.) 

Formula. 

Formulae will now be given for computing the quantities 
M m , B m , r\ m , M\, 0i, and yji for each kind of resistance separately. 
Their deduction will be given in the Appendix. If M , 6, or yj is 
desired, we can readily obtain it by observing that 

M = Mi + M m , 6 = 0i + dm, andYj = yji + Yj m . 
The following additional notation will be used: 
Let w = weight of reciprocating parts of valve gear in pounds. 

Wa^R 
Po = = throw of valve gear at the dead points in 

9 

pounds. 

Pi = total friction of valve and valve rod in pounds. 

P 2 = steam pressure on end of valve rod in pounds. 

P 3 = weight of valve, valve rod, and eccentric rod in pounds. 



192 DYNAMICS OF MACHINERY 

The formulae are as follows: 

1° Reciprocating Parts of Valve Gears. 

M m =^cos 6, B m =p -f-'sin 5, r> m = - p -ficos S, 
2 4 w p 2 a 8 w p 2 a 2 

M x =- ^f-'cos (2at + 5), d l = - p- -^sin (2 at + 5), 
2 4w p 2 a 

2° Steam Pressure on End of Valve Rod. 

M m =0, e m =-^^> y] to =0, 

W p 2 a 

Mi = — P 2 ri sin at, 0i = — ~ cos at, 

w p 2 a 

P2 gri . , 

TQi= Tism at. 

w p 2 a 2 

3° Weight of Valve, Valve Rod, and Eccentric Rod. 
M TO =0, e m =- — ^> Y) m =0, 

Mi=— Pzri sin arf, 0i = — ^ cos at 

w p*a 

P* gri . , 
w p 2 a 2 

4° Friction of Valve and Valve Rod. 

n , 2 Pin . , 
M m = sin 8 ; 

a _ ?JL T± \ 2 cos 3 + 5 sin 8 , 



w p 2 a f ir 



rim = - — ■ ■?! if- 2 - ^ + 5 2 ) sin 5 + 2 6 cos 8 I- 
ir w p 2 a 2 (\ 12 / ) 

When at<^-8: M 1= Ptf J - sin at ^?L_Y 

7T „ , , ~ / . . 2 sin 5\ 
When at>^-8: Mi= P^ifsin at J. 

tttl , ^ 7T „ „ Pi ori ( cos 8 + 5 sin 5 

When a* < - - 8: 0i = — V I ~ 2 

2 w p l a ( it 

. sin 5) 
+ cos at — 2 at > * 

7T \ 



GOVERNORS 193 



Tim. 4 ^ T * a p i ar i \ o • 5 cos 6 + 5 sin 5 

When at> s — 8: 0i = — ^- } 2 sin 6 — 2 

2 w p 2 a ^ T 

sin 5 



— cos erf — 2 atf 



7T 



wu 4 / x s jPi grri ( cos 5 + 5 sin 5 . . 

When erf < 7: — 5: yji = — ■£— -2 a/ h sni a/ 

2 w p 2 a 2 ^ 7r 

„,„ sin 5 ) 
— a 2 F > — Y) TO . 

7T ) 

When crf>£-S: 771 = — * -^U 2 (cos b + 5 sin 6) 

2 w p 2 a 2 ( 

— t sin 5 + 2 erf sin 6 

cos 5 + 5 sin 5 272 sin 5 J 
-2al sin at + err > — in m . 

7T IT ) 

5° Acfo'on 0/ Gravity on the Swinging Weight 
M m =0, m =^cos/3, ig m =-^sin/3, 

Mi = TFaa sin {at- ff), e 1 =-^ 1 cos (erf - 0), 

p z a 

TJi =- -^ sin (erf - j8). 

Corresponding Quantities for the Entire Resistance. 

The values of M m , 6 m , and yj to for the entire resistance are the 
algebraic sums of the corresponding quantities for the separate 
resistances. 

The values of Mi, d h and tqi for any given crank angle, for the 
entire resistance, are the algebraic sums of the corresponding 
quantities for the separate resistances. 

These formulae, therefore, enable us to plot at once any one of 
the resultant curves, whether the Mi, the d 1} or the tqi curve. 

By plotting the tq curve and the tq to line {KK') in Fig. 125, we 
can readily ascertain the greatest positive and the greatest nega- 
tive value of y]i, and by adding them without regard to sign we 
obtain the greatest angular travel of the swinging weight. 

Observe that the greatest positive and the greatest negative 
values of yji occur at crank angles for which 0i = 0. On the other 
hand, we can determine the greatest travel of the swinging weight 
that would be due to each kind of resistance separately by putting 
0i in each case equal to zero, solving for at, and substituting this 



194 DYNAMICS OF MACHINERY 

value in Yji; and, as there are two values of at, there will be one 
greatest positive and one greatest negative value. Then add the 
two numerically and we have the greatest travel for that kind of 
resistance. The results are: 



1° For reciprocating parts of valve gear, 



4 w p 2 a 2 



2° For steam pressure on end of valve rod, 2 — -^—5 ♦ 

w p L cc 

3° For weight of valve, valve rod, and eccentric rod, 2— -~^- 

w p l a l 

4° For friction of valve and valve rod: In this case the deter- 
mination will have to be made by trial, and the values substituted 
numerically. 



5° For the action of gravity on the swinging weight, 



2gx 



p 2 a 2 



By this means we can compare the effects of the separate resist- 
ances. 

In many cases, it will be found that the effect of the action of 
gravity on the swinging weight is greater than that of any of the 
others. 

Example. 

We will now apply these processes to the study of the shaft 
governor of the engine already referred to, i.e., a vertical single- 
cylinder engine, 9 inches by 8, rated at 34 horse power, under a 
steam pressure of 80 pounds per square inch gauge; the cylinder 
being above the crank shaft; the valve being a piston valve taking 
steam on the inside. The remainder of the general description 
has been already given. 

In the case of this engine we have 

1° Diameter of cylinder, 9 inches 0.75 foot. 

2° Stroke, 8 inches § foot. 

3° Rated H.P. at 350 r.p.m. and 80 pounds 

pressure and 0.3 cut-off 34 H.P. 

4° Apparent cut-off at 350 r.p.m 0.3 stroke. 

5° Weight of valve, valve rod, tension block, 

and check nut 14 . 65 pounds. 

6° Weight of rocker including trunion bear- 
ings, guides, etc 17.02 pounds. 

7° Excess weight of rocker arm 3 . 45 pounds. 

8° Weight of eccentric rod 10. 16 pounds. 

9° Weight of swinging weight 64 . 43 pounds. 

10° Weight moment of inertia of swinging 
weight about pivot, 4966 pounds- 
Cinches) 2 = 34.4861 lbs.-ft. 2 



GOVERNOKS 



195 



11° Distance of center of grav- ] 2 32 

ity of swinging weight 2 . 32 inches = — p=p feet, 
from pivot, ) 

12° Distance from shaft center ) 47incheg = 5_47 feet 
to pivot center, ) 12 

13° R = J4.60 inches = ^pfeet. 

14° Reciprocating ) = Q 16 3 = 2g ^ ^ 

weight J 

15° Effect of weight"] 

of valve and !■ = 14.65 + 3.45 - 10.16 = 7.94 pounds. 

valve gear J 

The dimensions of the governor and valve gear give: 



Apparent cut-offs.. 

£ in inches 

/8 in degrees 

R in inches 

5 in degrees 

. ( at cut-off / . . 
a I in degrees ) . . 



0.1 

2.170 
69.283 

0.845 
91.900 

36.750 



0.2 

2.140 

67.282 

0.865 

78.800 

53.000 



0.3 

2.120 
66.035 

0.910 
69.900 

66.500 



0.4 

2.090 
64.273 

0.975 
62.800 

78.250 



0.5 

2.060 
62.614 

1.065 
56.100 

90.000 



0.6 

2.020 
60.539 

1.190 
49.800 

101.750 



0.7 

1.960 
57.653 

1.395 
43.400 

113.500 



The friction of the valve and valve rod, etc., is assumed to be 
57 pounds. Only the results for 0.3 cut-off will be given here. 
They have been plotted in Fig. 125. 

The table on page 196 gives the values of M h 6 h and tji. 

The 2M curve has been plotted to a scale of 1 inch to 230 inch- 
pounds. 



Also, since — = %M, and as % = 0.0773, and as At is the 
At I I 



number of seconds required for ten degrees of crank angle, hence 
At = //o ^V = 2 to • Hence the number of radians per second rep- 
resented by any one ordinate of the 6 curve is found by multiply- 
ing the number of inches in that ordinate by (230) (0.0773) ^lo 

= 0.0852. As y) = =-zr , the number of radians represented by any 

one ordinate of the yj curve is found by multiplying the number of 

0852 
inches in that ordinate by * _ = 0.000405. 

J 210 



Dynamical Treatment of the Flywheel Governor with only One 

Swinging Weight. 

Suppose that the steam pressure is constant, and suppose that 
the engine is running under a constant load at a constant speed, 
barring the periodical oscillations due to the variation of the 
resistance already explained. 



196 



DYNAMICS OF MACHINERY 



RESULTS FOR 0.3 CUT-OFF IN INCHES OF ORDINATE 

FOR PLOT. 



Crank 
Angles. 
Degrees. 



M 1 





10 

20 

30 

40 

50 

60 

70 

80 

90 

100 

110 

120 

130 

140 

150 

160 

170 

180 

190 

200 

210 

220 

230 

240 

250 

260 

270 

280 

290 

300 

310 

320 

330 

340 

350 

360 



-1. 
-1. 
-1. 

+0. 
+0. 
+0. 
+0. 

+1. 
+1. 

+0. 
+0 
+0. 
+0 
-0 
-0 
-0 
-0 
-0 
-0 
-0 
-0 
+0 

+1 
+1. 
+1. 
+1 
+1 

+0 
+0 
-0 
-0 

-1 
-1 
-1 
-1 
-1 
-1 



569 
435 
276 
009 
444 
775 
934 
072 
019 
876 
655 
401 
129 
108 
290 
406 
455 
428 
365 
279 
206 
963 
254 
415 
434 
302 
057 
702 
273 
193 
623 
014 
324 
534 
683 
644 
569 



0i 



-2. 
-4. 
-5. 
-5. 
-5. 
-4. 
-4. 
-2. 
-2. 
-1. 
-0. 
+0. 
+0. 
+0. 
+0. 
-0. 
-0. 

-1. 
-1. 
-1. 

-2. 
-1. 
-0. 

+0. 

+2. 

+3. 

+4. 

+5. 

+6. 

+6 

+5 

+4 

+3 

+2 

+0 

-0 

-2 



541 

066 

451 

728 

525 

829 

073 

964 

038 

111 

367 

141 

383 

375 

.159 

.218 

.667 

.136 

.553 

.898 

.169 

.432 

.345 

.977 

.383 

.838 

.906 

.765 

.231 

.209 

.837 

.997 

.813 

.359 

.753 

.912 

541 



Vi 



+27.502 
+22.388 
+ 15.875 
+ 8.638 
+ 1.523 

- 5.490 
-11.532 
-15.808 
-18.789 
-20.521 
-21.102 
-20.745 
-19.715 
-18.295 
-16.244 
-14.283 
-13.075 
-12.222 
-11.788 
-11.704 
-11.983 
-12.136 
-11.537 
-10.436 

- 8.214 

- 4.344 
+ 0.509 
+ 6.003 
+ 11.844 
+17.619 
+22.899 
+27.275 
+30.896 
+32.995 
+32.903 
+31.063 
+27.502 



Moreover, we have crank angle at cut-off 
= 66° 30'. 

M m = 0.967" = 222.42 inch-pounds. 

6 m = 2.541" = 0.216 radians per second = 
12.37° per second. 

Vm =- 27.502"= - 0.0111 radians = 0.63°. 

Observe that in the preceding discussion 
one of the resistances has been omitted, 
on account of insufficient data, viz., the 
friction on the pivot P. The effect of 
this would be to somewhat dampen the 
oscillation. 

Observe also that the value of M m is the 
value which we should use for M 4 in the 
moment equation, and that the sum of 
M 4 and Sr 2 , the moment of the spring 
tension, would be equal to the moment 
of the centrifugal force at this cut-off, 
and hence equal to 



W 
9 



r*; 



and since W and £ are known a can be 
determined, i.e., the speed of the engine 
for that cut-off. 



Suppose that the load referred to above corresponds to a certain 
cut-off (as 0.5), and that at a certain instant a portion of the load 
is suddenly removed, so that the remainder corresponds to a 
shorter cut-off (as 0.3), and that we wish to ascertain all that 
occurs in consequence of the change of load. 

For the sake of simplicity in the discussion let us assume (a) 
that the change occurs immediately after cut-off, and (6) that any 
change in the compression before the next stroke has so little effect 
that we may disregard it. The first effect of the change is an un- 
balanced driving moment, and consequently an increase in the 
speed of the engine, and hence of the governor. 



GOVERNORS 197 

The amount of this increase of speed, by the time the next 
cut-off occurs, depends on the flywheel only, and not on the gover- 
nor, as the latter can produce no effect until the time of the next 
cut-off. 

The method by which the speed of the flywheel and hence also 
that of the governor at the time of the next cut-off is to be com- 
puted has already been explained under the heading " Acceleration 
of Flywheel when Load is Suddenly Changed," page 70. 

The method as applied to this case may be described as follows, 
viz.: 

Let a = speed before the change occurs, in radians per second. 
a = speed t seconds after change occurs, in radians per 

second. 
t\ = number of seconds elapsing before the next cut-off 

occurs after the change. 
cti = speed t\ seconds after change occurs, in radians per 

second, i.e., speed at time of next cut-off. 
/i = moment of inertia of flywheel (units being pounds and 
inches). 
M n = driving moment corresponding to original load, in inch- 
pounds. 
M q — driving moment corresponding to new load, in inch- 
pounds. 

Then if M denote the unbalanced moment in inch-pounds, we 
shall have at the instant when the change occurs M = M n — M q . 
If now we consider M to remain constant until the time of the 
next cut-off, we shall have 

- ■ -77 = M n - M q ; .*. a = a + j- (M n - M q ) t, 
g at i\ 

and hence 

«i = a + j- (Mn - M q ) h. . (1) 

From (1) we can determine the speed of the governor at the instant 
when the next cut-off occurs. 

We next need to determine the angular displacement which the 
swinging weight has undergone during the same interval, and there- 
fore the new angular displacement of the swinging weight (new 
value of i, and hence new value of £), and hence the new cut-off. 

For this purpose we must first ascertain the unbalanced moment 
acting on the swinging weight, which we will call M 8 . 

Let M e = unbalanced moment due to angular position. 
Mi = inertia moment of swinging weight. 
M d = moment due to dashpot if one is used. 

Then M. = M e + M » + M d . 



198 DYNAMICS OF MACHINERY 

In this discussion we will assume that there is no dashpot, therefore 

that M d = 0. Then we shall have 

M 8 = M e + Mi. 

To find M e graphically, refer to Fig. 118. Draw a ray through 
the point on the M f curve corresponding to the original cut-off (0.5) 
and take the difference between its ordinate at the new cut-off 
(0.3) and the ordinate of the M f curve at the new cut-off (0.3). 
This difference will represent M e to scale, for the ordinate of the 
ray at the new cut-off (0.3) is the moment of the centrifugal force 
corresponding to the new cut-off and the original speed. The 
value of M e can be calculated analytically instead of being de- 
termined graphically, by computing each of the ordinates men- 
tioned, and subtracting. 

To determine M h proceed as follows (Fig. 117): 

Let r = OG. 

p = perpendicular from P to a line drawn through G at 

right angles to the line OG. 
Io = moment of inertia of swinging weight about G. 

Since the angular acceleration of the flywheel, and hence of 

da 
the governor, is-^r, the linear acceleration of its center of gravity 

da 
G is r -jt , though in the opposite direction. Hence the accelerat- 
ing force at the center of gravity is 



W da 
9 dt 



and its moment about P is 



W da 

r oP ~17 • 

g dt 

The swinging weight, however, has also an inertia moment due 
to its rotation about its own center of gravity, and this is equal to 

7o da 
g dt 
Hence the inertia moment of the swinging weight is 

n/r _ (Io + Wr p) da 

Mi — -J7 * 

g at 

Moreover, substituting for -^ its value j- (Mn — M g ), we obtain 
Mi= I JL ±WM> (Mn l _ Mq)> 






GOVERNORS 199 

and hence 

M, =M e + h \ WroP (M n - M t ). 

In making this addition, however, care must be taken to see whether 
the last term is essentially negative or not with regard to M e . 
Having thus found M s , we may proceed as follows : 

Let J = moment of inertia of the swinging weight about P. 
O = angular velocity, in radians per second, of swinging 

weight at the instant when the change occurs. 
Y]o = displacement, in radians, from an arbitrary position, 

of the swinging weight at the instant when the 

change occurs. 
= angular velocity, in radians per second, of swinging 

weight, t seconds after change occurs. 
tq = displacement, in radians, from the same arbitrary 

position, of the swinging weight, t seconds after 

change occurs. 
0i = velocity, in radians per second, of swinging weight at 

the time of the next cut-off. 
tqi = displacement, in radians, from the same arbitrary 

position, of the swinging weight at the time of the 

next cut-off. 



Then we shall have 

dh^ __ 

dt 2 ~ ^" " dt 






dt 



+ 1 M s t, = O + | M 8 t, 0! = O + | M&. 



Hence 



TQ = K M^ | 2 + d t + Y) ; /. IJi = (| M)j ^ + O *! + Y] . 

Having thus ascertained the speed and the position of the swinging 
weight at the time of the new cut-off, i.e., the speed and the per 
3ent of cut-off when the first new cut-off occurs, we start again 
with new values of M n , O , and tq , and proceed in a similar manner 
to the next stroke, and so continue, until the speed and the cut-off 
reach simultaneously the values corresponding to the new load. 

This is of course a very slow process, but the complete work- 
ing out of the problem, by integrating, and eliminating between 
the differential equations 

L*42-M.-M, and *-% - M„ 

g dt g dt 

taking account of the variation of M n and taking account of the 
presence of a dashpot, is a very complicated process, and will not 
be given here. 



CHAPTER V. 

BODIES WITH A HIGH ROTATIVE SPEED. 

Moving Axes. 
In the Case of a Moving Body, with One Point Fixed in Position 

If, as is often the case, we find it convenient to refer the motion 
to a set of rectangular axes fixed in the body, and all passing 
through the fixed point (this being a set of moving axes), it will 
be necessary to have also a set of axes fixed in space, to which can 
be referred the motion of the moving axes. 

In such cases, the motion of any point in the body will be the 
resultant of its motion with reference to the moving axes (i.e., 
those fixed in the body), and of the motion of the system of moving, 
axes with reference to those fixed in space. 




Fig. 126. 

The equations which express the relations between the com- 
ponents of the angular velocities about the moving axes, at a given 
instant, and the position and motion of the body at the same 
instant, are known as Euler's Kinematical or Geometrical Equa- 
tions. To deduce them, proceed as follows, viz.: Let be the 
fixed point. About as a center describe a sphere, with radius 
equal to unity. Let OX, OY, OZ be the axes fixed in space. 
OX = OY = OZ = 1. Let OX', OY' , OZ' be the positions of 
the moving axes at any given instant, as at the end of time, t y 
after starting, and let OX' = OY' = OZ' = 1. 

200 



BODIES WITH A HIGH ROTATIVE SPEED 201 

In order to fix the position of the body at that instant, and 
hence that of the moving axes OX', OY', OZ' with reference to the 
fixed axes OX, OY, OZ, we need to know the values of the three 
angles ZOZ' = 0, XON = 4,, and NOX[ = <j>, where NON' (called 
the line of nodes) is the line of intersection of the plane through 0, 
at right angles to OZ' , with the plane XOY. Another mode of 
conceiving these angles is to consider the motions needed to 
transfer the moving axes from coincidence with the fixed axes 
to their actual position. These motions may be regarded as 
having been produced by tipping the horizontal plane passing 
through 0, and fixed in the body, around the line NON' (called 
the line of nodes, N and N' being the nodes, and the angle XON 
being denoted by rp), through an angle ZOZ' = 6, and then turning 
the body around OZ' , through an angle </>, so that NOX'= EOY' 

By adopting this notation for these angles, we obtain: 

dd 
1° Angular velocity of the body about ON = -=-> represented 



graphically by a vector along ON. 
ngular velocity of ON aboul 
ically by a vector along OZ. 



/7 / 

2° Angular velocity of ON about OZ = -j- > represented graph- 



3° Angular velocity of the body about OE = sin 6 -j- » repre- 
sented graphically by a vector along OE. 
4° Angular velocity of the body about OZ' , relatively to line of 

nodes = -rr» represented graphically by a vector along OZ'. 

The first, second, and fourth are evident at once. To prove the 
third, observe that -r- is the angular velocity of Z' about OZ; the 
linear velocity of Z' in a horizontal direction will be consequently 
Z'G -77 = sin 0-j- , since Z'G = sin 0, and since OE is perpendicular 
to OZ', and OZ' = 1. Hence we have 

sin 6-^ 

Angular velocity about OE = nvf = sin B-j-- Q.E.D. 

(JZj dt 

On the other hand, at the instant in question, the body is revolving 
about some instantaneous axis which passes through (this axis is 
not shown in the figure), with an angular velocity which we will 
call co, and we will call the components of co about OX', OY', OZ' 
respectively, cci, co 2 , co 3 . Hence we have, 

5° coi = angular velocity about OX', represented graphically by 
a vector along OX'. 



202 DYNAMICS OF MACHINERY 

6° co 2 = angular velocity about OY', represented graphically by 

a vector along OY'. 
7° co 3 = angular velocity about OZ' ', represented graphically by 

a vector along OZ'. 

Resolving a>i and <a 2 along ON and OE, and equating the alge- 

d9 
braic sum of their components along ON to -77 , we have 

rift 

=£ = coi cos X'OiV + o) 2 cos FW, 
or -17 = 0)1 cos — o) 2 sin 0, (1) 

since XW = 0, FW = ^+ 0. 

Equating the algebraic sum of their components along OE to 

sin -r- f . 
at 

we have sin 6 -7- = «i cos X'OE + o) 2 cos Y'OE, 

d\l/ 
or sin 0-7- = on sin + o) 2 cos 0, (2) 

since X'OE = | - 0, F'0# = 0. 

Solving equations (1) and (2) for 0)1 and co 2 , we obtain 

dd , , d\[/ . . . , . 

0)1 = -7: cos + -77 sm sin (3) 

d0 . . dyl . n ... 

o>2 = — -77- sin + -7- sin cos (4) 

Moreover, to obtain 0)3, observe that the angular velocity of the 
body about OZ' is made up of two parts, viz., (a) the component 

along OZ' of the angular velocity -7- of the line of nodes about OZ, 
or -^ cos 0; and (6) the angular velocity of the body about OZ' 
relatively to the line of nodes, or -7- • 

Hence we have 

W3= _ cos() + _ (5) 

Equations (1), (2), (3), (4), and (5) express the relations sought 
for and needed. 



BODIES WITH A HIGH ROTATIVE SPEED 



203 



Euler's Dynamical Equations. 

Assume a body in motion, with one point fixed in position. 
Let the moving axes be the principal axes through 0. Let A, B, 
C be the moments of inertia of the body about the moving axes. 
.Let L, M, N be the moments, about the moving axes, of the 
forces acting on the body. 

Let coi, a>2, C03 be the angular veloci- 
ties, in radians per second of the body 
about the moving axes. 

Consider the body at the instant 

when its position is such that the 

moving axes coincide with the fixed axis 

OX, OY, OZ. Let P be any point in 

w 
the body. Let m = — = the mass of 

9 
an elementary volume at P. Let x, y, z 

be the coordinates of P. Then we 

shall have, 

dz 



»; 



f~ 



— X 



Fig. 127. 



dx 
dt' 


dy f 

dt ' 


d 2 x 
dt 2 ' 


d 2 y 
~dt 2 



-£> -j- will be the linear velocities of P, in directions 



dt 



parallel to OX, OY, OZ. 



d?z 
-tz 1 -£>> ~r^ will be the linear accelerations of P, in direc- 
dt 2 dt 2 dt 2 

tions parallel to OX, OY, OZ. 

A perusal of the figure will give the following six equations, viz. : 



dx 
~dt 
dy 
dt 
dz 



= 2C0 2 — 2/C0 3 . 



= xo)s — zooi. 



(i) 

(2) 
(3) 



d 2 z 
L = 2m [y-^ 



d 2 x 



dt 2 

*r ~ / d2 y 



d 2 y 
H 2 

d 2 z 

M = Sm ( z ^ — x -Tp 

d 2 x 
d? 



y 



)• 



(4) 

(5) 
(6) 



Differentiating (1), (2), and (.3), substituting the resulting values 
in (4), (5) : and (6), and observing that, for principal axes, 



we obtain, 



Xmyz =2mxz = Xmxy = 0, 

d 2 x dz dwo dy dwz 

—r 2 = o} 2 —,-\- z^ T - cos ^7 - y -jt^ 



di 

d 2 y 



dt 



dt 



dt 



dt 

do)\ 



dx c?co 3 dz 

dp ~ U3 Jt + x ~di ~ ai li ~ z dt 



d 2 z dy dcoi dx 

= 0)1— + y-rr - co 2 



dt 2 



dt 



dt 



dt 



x 



dd02 

~dt' 



204 DYNAMICS OF MACHINERY 

d 2 x 9 9 11 ^co 2 doi-i 

-j- 2 = ywi(j0 2 ~ XO) 2 Z ~ XW 3 2 + 2C0iC0 3 + Z—r- — y —r- , 

d 2 y dcos dtoi 

~TT 2 = ZW2W2 — yO)3 Z — yCQr 2 + XCO1CO2 + x -TT — z -J-, 

d 2 z 00, ■ dcoi du<> 

^ = XCO1CO3 — 2C0i 2 — 2C0 2 2 + 2/0) 2 C03 + y -TT — X -j- ■ 

^ 2m (z 2 + y 2 ) + <o 2 co 3 2m (?/ 2 - z 2 ) = L . . . (7) 

^- 2 2m (z 2 + x 2 ) -f W3 coi2m (z 2 - x 2 ) = M . . (8) 

^ 2m (a: 2 + ?/ 2 ) + co!co 2 2m (x 2 - y 2 ) = N. . . (9) 
These equations readily reduce to the following, viz. : 

A ^ft + (C " B) ***** = Lg (10) 

B lH + (A ~ C) C ° 3C ° 1 = Mg ' ' • m (11) 
C^+(B-A)^ 2 = Ng. . . . (12) 

These three are Euler's dynamical equations, and enable us to 
express the moments of the forces about the moving axes, in terms 
of the angular velocities about the same axes. 



SYMMETRICAL TOP. 

Assume a moving body, Fig. 128, symmetrical about OZ r , with 
one point fixed in position. To find the position of the body 
at any given instant, and to study its motion. 

Let OX, OY, OZ be the fixed axes (OZ being vertical). 

OX', OY', OZ' be the principal axes, fixed in the body, and let 
their position at any instant be the position of the moving 
axes at that instant (the lines of nodes being NON'). 

A, B, C be the principal moments of inertia, at 0, of the body 
and let B = A. 

ZOZ' = 6, XON = f, and NOX' = <j> (all in radians). 

do be the initial value of 6. 

wi, co 2 , W3 be the angular velocities, in radians per second, of 

the body about the moving axes. 

E 2 

jr— be the initial actual energy of the body, i.e., that when 6 = O » 

A 9 



BODIES WITH A HIGH ROTATIVE SPEED 



205 



— z be the initial angular momentum of the body about OZ, 

i.e., that when = O . 

W be the weight of the body. 

h = OG = distance from to G, the center of gravity of the 
body (G being on the line OZ'). 

M r = moment of resultant couple acting on the body. 




Fig. 128. 

We then have, that the force of gravity acting at G, and the 
vertical component of the reaction at 0, form a statical couple, 
in the plane ZOZ' , whose moment is 

M r = Wh sin (1) 

The components of M T about OX', OY', and OZ' are respectively 

L = Wh sin cos . . (2) N = (4) 

M = - Wh sin 6 sin 0. . (3) 

Also, when ZOZ' changes from O to 0, the work performed by the 
couple is 



Jr*d r*9 

M r dd = Wh I sin Odd = Wh (cos O - cos 0). 
0Q *J 0q 



(5) 



Substitute now the values of L, M, and N, given above, in the 
Euler dynamical equations, and we obtain 

A -^ + (C - A) o) 2 co 3 = Wgh sin cos </>. . . (6) 



A ii +(A -o^ 2 = 



Wgh sin sin </>. . (7) 

(8) 

From (8) we obtain o> 3 = a constant. Let n = angular velocity of 



dt 



206 DYNAMICS OF MACHINERY 

spin, in radians per second, about OZ'. Then we have m = n. 
Therefore equations (6), (7), and (8) become 

A -^ + (C - A) no> 2 = Wgh sin cos </>. . . (9) 

A -^ + (A - C) no)i = - Wgh sin sin 0. . (10) 

co 3 = n (11) 

These are the fundamental equations, which enable us to solve 
problems concerning the top. 

In certain special cases they are also the forms most convenient 
for use, whereas, in other cases, it will be necessary to employ two 
other equations, which may be called respectively the actual- 
energy equation and the angular-momentum equation. 

While these can be derived from (9), (10), and (11), they can 
also be obtained directly, in an easier manner, as follows : 

(a) The actual energy at any given instant (when ZOZ' = 0) is 

Aon 2 AW Cn? 
2<7 + 2g ^ 2g' 

and this is equal to the initial actual energy plus the work done 
by gravity. Hence we have 



W Ao><? Cn 2 # 2 

-s r- -^ h 75 — = 75 h Tr /i (cos O - cos 0). . (12) 

2g 2g 2g 2g 

(6) As gravity, the only force acting, is vertical, it cannot 

change the angular momentum about OZ. Hence, this angular 

M z 
momentum must be constant, and equal to its initial value — -- 

g 

To obtain the expression for this angular momentum about OZ 
and hence to obtain the angular-momentum equation, proceed as 
follows, viz.: 

The angular momenta about the moving axes OX', OY', OZ' 

are respectively 

Aco\ A0J2 1 Cn 
f f and 

99 g 

Resolve the two first into components along OE and ON. Those 
along ON do not affect the angular momentum about OZ. The 
sum of the components along OE is 

— cos X'OE + — cos Y'OE = - («i sin + co 2 cos 0), . (13) 
g g 9 . 

and, if this be resolved, along and perpendicular to OZ, its com- 
ponent along OZ is 

A 

— (coi sin <b + co 2 cos 6) sin 6, 

g 

\71 ( 71 

whereas the component of — along OZ is — cos 6. 



BODIES WITH A HIGH ROTATIVE SPEED 207 

Hence we have 

— (coi sin 6 + 0)2 cos d>) sin 6 -\ cos = — -, . (14) 

g 9 g 

and this may be called the angular-momentum equation. Equa- 
tions (9), (10), (11), (12), and (14) may be called the fundamental 
equations, some of which are needed in solving any problem 
connected with the action of the top. Nevertheless, considerable 
changes in the form of some of them are often desirable. Inas- 
much as, in many cases, the position of the body at any instant is 
called for, and hence the corresponding values of 0, </>, and \j/, it 
will be convenient to substitute for coi and co 2 in equations (11), 
(12), and (14), their values in terms of 0, </>, and \f/, as given in the 
Euler kinematical equations. We thus obtain in place of (9), 
(10), (11), (12), and (14) the following: 

A-jr+ (C — A)nca 2 = Wgh&mdcos </>. . . . (15) 

A ^ + (A - C) w«i = - Wgh sin sin <f>. . . (16) 

d -±=n-^cos6 (17) 

at at 

2 , . , „ /#Y E 2 + 2 Wgh cos 6 - Crfi 

+ Sm ° \W = A 

-^cosO (18) 

SPECIAL CASES. 

The above is the general theory. In order to solve examples 
under it, we should need, either to impose certain conditions, or to 
adopt certain approximations that would result in simplifying the 
equations. We are generally concerned, however, with special 
cases, and will now consider some of them. 

FIRST SPECIAL CASE. 

The only condition imposed in this case is that the initial spin 
is around the axis OZ' . This is the most general of the special 
cases that will be considered here. 

By imposing the above-stated condition we obtain 
E 2 = Cn 2 and M z = Cn cos . 
Making these substitutions in equations (18) and (19), they become 

Jt) = A (C0S d ° ~ C0S 6) ~ Sm2 e \Jt) ' * (20) 



c 



sin 



d(^\ = ^j (cos So - cos 6) (21) 



208 DYNAMICS OF MACHINERY 

Substituting the value of -^from (21) in (20), and reducing, we have 

^j = (cos O - cos 6)\2 WgAh sin 2 

- C 2 n 2 (cos do - cos 0) \ . . . . (22) 

dO 
To find the maximum and minimum values of 0, put -r = 0. That 

is find the values of that satisfy the equation 

(cos 0o - cos 6)\2 WgAh sin 2 - C 2 n 2 (cos O - cos $)\ = 0. 

This equation will be satisfied by three values of 0, which we will 
call respectively, 0i, 2 , and 03. 

Evidently one of these values, which we will call 2 , is equal to O , 
while the other two, viz., 0i and 03, must satisfy the equation 

2 WgAh sin 2 - C 2 n 2 (cos O - cos 0) = (23) 

If now we write z = h cos = vertical height of G, the center of 
gravity above 0, and hence z = h cos O , Z\ = h cos 0i, z 2 = h cos 2 , 
and Zz = h cos 3 , then equation (23) may be reduced to 

C 2 n 2 C 2 n 2 

z2 -2WJA Z - h2+ 2WJA ZQ = 0; ' * ' (24) 

.. , , C 2 n 2 

or if we let a = 



2TF^A 

it may be written 

z 2 - az - h 2 + az = (25) 

If for z in the first member of (25) we put oo, the result is a plus 
quantity, if either z or h, & minus quantity, and if — oo, a plus 
quantity. Therefore, one of the roots is greater than h, and the 
other is less than z . 

Solving the equation, we obtain z\ and z%, and as z 2 = z , we have 



a I o? d /a 2 

Zl = 2 ~ V T~^" h2 ~ aZ °' Z2= Zo > 2s = 2^V4^^~ aZ °' 

Of these the last is greater than h, hence we are only concerned with 
the first two, viz., z\ = h cos 0i, and z 2 = h cos 2 = h cos O . Ob- 
serve that, in order that z may be positive, we must have 

az»>h 2 - ' n* > 2 AWgh 
az Q >h, .. n > C2cos o 

The precessional velocity may be found from equation (21). It is 

d\p _Cn cos O — cos . _,. 

It ~ A sin 2 { } 

When = 0o the precessional velocity is zero, and then the axis 



BODIES WITH A HIGH ROTATIVE SPEED 



209 



OZ' is an element of the right circular cone whose half-angle is O . 
As increases, and hence as z = h cos decreases, the precessional 
velocity increases until = 0i, when it has its greatest value, and 
then the axis OZ' is an element of the right circular cone whose 
half-angle is 0i. 

After -j- has been found -^ can be found from equation (17). 




Solution of Numerical Examples under the 
First Special Case. 

To put the equations in a more con- 
venient form, for the solution of numerical 
examples, we will make use of the fol- 
lowing approximations. Assume n to be 
very large, then both 0i — O and — O 
will be small. Write — O = (3 where /5 
is small, and consequently we will write 
cos (3 = 1 and sin (3 = /3. Hence we shall 
have Fl s- 129 ' 

= 0o + |8. 
dd _ d@^ 
dt ~ dt 
sin = sin (0 O + jS) = sin O + (3 cos O nearly, 
cos = cos (0 O + j8) = cos O — jS sin O nearly, 
cos O — cos = )8 sin O nearly. 

j8 2 sin 2 0o 

— . 2 = |3 2 nearly. 
sin 2 *^ 

Dividing out equation (22) by A 2 sin 2 and making these substitu- 
tions, we have 

'dj8\ 2 C 2 n 2 ( /WgAh . a \ a M ) 

ft)=^r{w sme r-^\- • • (27) 

or if we let e = J? 9 9 sin O (28) 

C 2 n 2 

(27) becomes 

dfi Cn , 

^ = T V2^^ 2 (29) 

Hence 

dt = £-=M== (30) 

Cn V2 e(3 - (3 2 

Hence integrating (30), we have 

. A . -V^N . R . (Cnt\ f . 

t = ~- versin (-); .. /3 = e versml — t-1. . (31) 



210 



DYNAMICS OF MACHINERY 



The corresponding values of -~ > -n » ^, and are as follows : 



dt~ Cn l l \AJ 

, WgJi WghA (Cr 



A )' 



d<j> 



= W 



Wgh 



dt 

<f> = ]n — 



cos 



cos 0o ] 1 



! 



©i- • 



COS O > £ + 



sin - 



Cn —""J" ' C 2 n 2 
Observe that these equations do not hold unless n 2 > 



(32) 
(33) 
(34) 

(35) 



\A 

2 WghA 

C 2 cos do 7 
and that the greater n, the closer the approximation. 

Since 0i is the greatest value of 0, and since the greatest value of 



versm 



(Cnt\ 



Cut 
is 2, and occurs when— 7- = 71-, equation (31) gives 



A 

0i = O + 2 e. 



(36) 



7T 



Moreover, t = -~- is the time in seconds of one-half of a period, i.e., 

the time required for the axis OZ r to go from one of the limiting 
right circular cones to the other. Hence if T is the time in seconds 
of one period, 

T = 2 -£ < 37 > 

Moreover, if \f/i denote the precessional angle of one period, it can 

9 A 
be found by substituting — ^ — for t in equation (33). Hence 



2 Wgh 1 



(38) 



To solve a numerical example under this case proceed as follows : 

1°, find e from (28); 2°, find 6 l from 
(36); 3°, find T from (37); and 4°, find 
iMrom (38). 

Example I. — Let the top consist of 
a steel ring with a square section spin- 
ning about 0, at 2000 revolutions per 
minute. Take units in inches and 
pounds. Outside diameter = 4". In- 
side diameter = 3". Thickness, J". 
OG = 1". Weight of metal = 0.28 
= 5°. n = 209.44 radians per second. 
B = 0.9884. C = 1.2031. 



f 



Fig. 130. 

pound per cubic inch. O 
W = 0.385 pound. A 



BODIES WITH A HIGH ROTATIVE SPEED 211 

Then we have 

e = 0.00022 radian = 0.126° = 0.75' = 45". 
2 e = 0.00044 radian = 0.252° = 1.5'. 
0! = 5° 1' 30". T = 0.0247 second. 
fa = 0.0047 radian = 0.269° = 0° 16' 8". 

Example II. — Assume a hydro-extractor where, the units being 
feet and pounds: W = 378.54 pounds. A = B = 3775.21. C = 
449.62. h = 2.8 feet. Revolutions per minute, 4000. n = 418.88 
radians per second. O = 5°. 

Then we have 

2 e = 0.00064 radian = 0° 2' 10". 0i = 5° 2' 10". 
T = 0.0132 second fa = 1° 18' 23". 

SECOND SPECIAL CASE. 

To determine the conditions which will cause the axis OZ' to 
move in the surface of a right circular cone, so that 6 shall be 
constant, and equal to 0o throughout, and to study the motion. 

A top under these conditions is said to be in steady motion. 
In this case we will not assume that the initial rotation takes place 
about OZ'. 

Let 1? Q 2 , and n be the initial values of «i, a> 2 , and 003 re- 
spectively. Starting with the general equations (15), (16), (17), 
(18), and (19), we shall have in this case that 

E 2 = A (O1 2 + « 2 2 ) + Cn\ 

M z = A VQi 2 + 2 2 sin O + Cn cos O , 

* = *„; , g-0. 

The general equations then reduce to the following : 

(15) becomes A -=— + (C — A) nut — Wgh sin O cos 0. . (39) 

(16) becomes A -r- 2 + (A — C) nu\ = — Wgh sin O sin <f>. (40) 

(17) becomes -^ = n — -£ cos O (41) 

(18) and (19) both' reduce to 

$// VOi 2 + o 2 2 

-rr = : — 7 = a constant = a (say). . . (42) 

dt sin O 

We thus have 

dS ^ d-d/ deb 

e = e °> dt = ' Tt= a > * = ai ' Tt = n - aCose, » 

4> = (n — a cos O ) t. 



212 DYNAMICS OF MACHINERY 

In order, now, to derive the conditions that will produce this result, 
we need to make use of equations (39) and (40). In order to do 
so, however, we must first express coi and co 2 in terms of 0, <j>, and \p, 
by means of the Euler geometrical equations (3) and (4), page 
202. Putting 

dd _ d\b , , n N , 

°' Tt = ' ~Ht = a ' a <f> = {n — a cos do) t 

in these equations, we obtain 

coi = a sin O sin in — a cos O ) t, and co 2 = a sin O cos (n — a cos do) t. 

By differentiation we have 

-T7 = (na — a 2 cos d ) sin 6 cos (n — a cos $o) t> 

-TT = — (na — a 2 cos do) sin O sin (w — a cos O ) ^. 

Hence equations (39) and (40) become respectively 

— sin O \Aa 2 cos O — Cna + Whg\ cos (n — a cos O ) ^ = 0, 
sin O \Aa 2 cos O — Cna + W%| sin (n — a cos O ) t = 0. 

As these two equations hold for all the values of t, we must have 

Aa 2 cos O - Cna + Whg = 0, ... (43) 

and this is the condition for steady motion. 

Before discussing the significance of equation (43), we will first 
summarize the results as follows: 

Wh= \(-)n -(-) a cos d i a. . . . (44) 

# = a = Vfli" + 2 2 cosec O (45) 

^ = at (46) 

-j- = n — a cos O . (47) 

= (n — a cos do) t (48) 

Discussion of Equation (43) or (44). 
I. Multiply equation (44) throughout by sin O , and we have 

Wh sin O = \ {-) n sin O — [ — ) a sin O cos d \a. . (49) 

t\g/ \g/ ) 

Consider now the meaning of the separate terms in this equation. 

1° The first term Wh sin O is the moment of the force acting on 
the top, as explained in equation (1). It is called the Torque, and 
will be denoted by T. Its axis is the line of nodes, ON., in Fig. 128. 



BODIES WITH A HIGH ROTATIVE SPEED 213 

2° The first term in the brackets (— J n sin O is the component 

of (— ) n (the angular momentum about OZ' ', Fig. 128, along a 
line at right angles to OZ and to the line of nodes. 

3° The angular momentum about OE, Fig. 128, as was shown 
in expression (13), is (f) (c, sin * + * 2 cos ,), which, in this case, 

reduces to ( — J a sin O . The component of this angular momentum 
along the line at right angles to OZ and to the line of nodes is, 

therefore, ( — J a sin O cos O . 

These two angular momenta, viz., 

{ — jn sin O and ( — J a sin O cos O , 

riave opposite signs, hence if we call the resultant angular momen- 
tum about the line at right angles to OZ and to the line of nodes 
M e , we shall have 

M e = \ (— )n sin O — [ — ) <* sin o cos O £ • 

(\g/ \g/ ) 

Equation (49) expresses the fact that the torque is found by 
multiplying M e by the precessional velocity a, or 

T = M e a . (50) 

When the torque axis momentarily coincides with OY, then the 
M e axis coincides with OX. 

Observe that, in order to maintain the precessional velocity a, 
a torque is necessary, whose magnitude is T = M e a, and whose 
axis is at right angles to the axes of M e and a. 

Hence the application of a torque about OX causes the top to 
precess about OZ. 

II. Equation (42) gives us 

Vo^ + n 2 2 



a = 



sin O 



Hence, O^ and n being given, to maintain a steady motion, the 
smaller the angle O , and hence the greater the value of z = h cos O 
(i.e., the higher the center of gravity), the greater the precessional 
velocity required. 

Moreover, consider the action of the top assumed in the first 
special case, which rises and falls periodically. When the center 
of gravity has reached its highest position the precessional velocity 
has decreased to zero. Then the top begins to fall; as it falls, 



214 



DYNAMICS OF MACHINERY 



that portion of the angular momentum about OZ that is due to 
the spin, i.e., (— J n cos 0, decreases, and, since the total angular 

momentum about OZ cannot change, it follows that that portion 

that is due to the precession is 
increased. This action goes on 
until the lowest position of the 
center of gravity is reached, when 
the precessional velocity has 
reached its greatest value, and 
then the center of gravity begins 
to rise, these phenomena being 
repeated periodically. 

Again assume a gyroscope with 
its axis inclined upwards as shown 
in the figure. One way of increas- 
ing the precessional velocity is 
by increasing the torque, say by 
adding a weight at X; this causes 
a tendency to dip, and this tend- 
ency results in increasing the pre- 
cessional velocity. Hence, work 
is done on the top, and hence, 

as its kinetic energy is not much increased, its potential energy 

is increased; thus the center of gravity rises. In view of all the 

above, we may say: 

(a) Hurry the precession, the top or gyroscope rises; 

(b) Retard the precession, the top or gyroscope falls. 

III. Solving (43) for a, we obtain 

Cn db VC 2 n 2 - 4 WhgA cos O 




Fig. 131. 



a = 



2 A cos B< 



When cos is negative, the values of a are always real. When 
cos is positive, then, in order that the values of a may be real, 
we must have 

C 2 n 2 > 4 WhgA cos O ; 

4 WhgA cos O 



n 2 > 



O 



/ THIRD SPECIAL CASE. 



In this case we will assume steady motion, but we will impose 
the additional condition that the axis OZ' revolves in a horizontal 
plane; hence that 



= O = 



TV 



cos O = 0, sin O = 1. 



BODIES WITH A HIGH ROTATIVE SPEED 215 

Imposing these conditions, we have 

E 2 = A (Oi 2 + 2 2) + Cn 2 . 
M z = A ViV + Oa 2 . 

ft - ft - w - • de - n 

The general equations then reduce to the following : 

(15) becomes A -j- + (C — A) nco 2 = Whg cos <f>. . (51) 

(16) becomes A ^ + {A - C) rua x = - Whg sin 0. . (52) 

(17) becomes — = n (53) 

(18) and (19) both reduce to 

Jj = V^x 2 + 2 2 = a constant = a (say). . . (54) 
We thus have 

d = d ° = 2' -dt = > dt =a > + = at > Tt =n > * = nt 

Hence equations (51) and (52) become respectively 

(Whg — Cna) cos nt = 0, and (Whg — Cna) sin n£ = 0. 
As these two equations hold for all values of t, we must have 

Wh = (-)na (55) 

Summarizing the results, we have 

Wh = (-)na (56) 

# = a = VoTTo? (57) 

^ = at (58) 

75-» <«» 

= nt (60) 

Discussion of Equation (55) . 
In this case it is evident that T = Wh, and that M e = (-)n. 
Hence T = M e a 



216 DYNAMICS OF MACHINERY 

as in the preceding case. Hence we have 

(61) 



r -0- 



and this gives the torque required to maintain the precessional 
velocity a, and shows that a torque about OY (say) results in a 
precessional velocity a about OZ where 

T 
<*=77n- (62) 



© 



n 



Application of the Principles of the Gyroscope in Engineering. 

The following is a list of some of the applications of the principles 
of the gyroscope to Engineering Problems: 

1° Balancing parts of machinery which have a high rotative 

speed. 
2° Steering of torpedoes. 
3° Steadying of vessels at sea. 
4° Brennan's monorail car. 
5° Gyroscopic compass. 

It is not the purpose of this treatise to give a detailed description 
of the manner of making each of these applications, but merely 
to point out how the principles already developed are employed 
in their solution. 

1° Balancing parts of machinery which have a high rotative speed, 
such as hydro-extractors, centrifugals, steam turbines, dynamo ar- 
matures, etc. — The meanings of standing and of running balance 
have been stated already, and it has been explained that before 
attempting to obtain a running we should first secure a standing 
balance. After a standing balance has been secured, the rotating 
piece to be balanced is usually mounted in a manner, as nearly as 
is feasible, similar to that in which it is to be used, and it is then 
run at a speed, as nearly as may be, identical with that at which 
it is to run in practice. If it is not in running balance, we need 
to ascertain the plane, and the sense, of the disturbing centrifugal 
couple. When this is known, the amount of each of the two 
weights to be added, to counterbalance the disturbing couple, can 
be, and generally is, determined by successive trials. 

As to the position of the plane of the unbalanced couple, it is 
easily determined when the speed is low, as in card cylinders, 
whose speed is seldom more than 200 revolutions per minute. 
For this purpose, the card cylinder, after a standing balance has 
been secured, is placed (Fig. 87) in a horizontal position on flexible 
bearings, which, when the cylinder is not in running balance, 
oscillate horizontally. Then chalk marks, made on the portions 



BODIES WITH A HIGH ROTATIVE SPEED 217 

that run out, determine the plane and sense of the disturbing 
centrifugal couple. This condition would be commonly expressed 
by saying that it runs out on the heavy sides, so that the counter- 
weights should be placed on the opposite or light sides. 

In the case of the hydro-extractor, which runs at a high speed, on 
a vertical shaft supported and driven from below, and resting on a 
flexible bearing, we have an arrangement similar to a top, except 
that the resistance of the rubber, and the friction in the bearing, 
and the influence of the driving belt, modify somewhat the motion. 
Were these resistances absent, and were the hydro-extractor in 
running as well as in standing balance, its motion could be pre- 
dicted from the preceding discussion. 

Inasmuch, however, as, when it is only in standing and not 
in running balance, it will run out, it is easy to determine the part 
that runs out by chalking it, but after this has been done the 
position of the plane of the disturbing centrifugal couple depends 
upon the speed. Thus at low speeds it runs out on the heavy 
side, at very high speeds on the light side, and at intermediate 
speeds the angle made by the plane in which it runs out, and the 
plane of the disturbing couple, varies with the speed. Hence we 
must either perform the experiment at very low or at very high 
speeds, or, if we wish to perform it at the speed at which it is to 
run in practice, we must previously determine, by experimental 
investigation on a similar apparatus in perfect balance, the angle 
between the two planes, when we introduce purposely a definite 
disturbing couple. When the apparatus has been put in running 
balance, we can predict its behavior approximately from the previ- 
ous discussion. 

In the case of a centrifugal machine, which runs at high speed 
on a vertical shaft suspended and driven from above, and sup- 
ported in a flexible bearing, we have, as it were, an inverted top, 
the motion of which is modified by the resistances at the bearing 
and by the influence of the driving belt. 

In both of these cases, when once the plane of and the sense 
of the disturbing centrifugal couple has been ascertained, weights 
are added in such a way as to introduce an equal and opposite 
couple. 

2° Steering a Torpedo. — There are several devices for auto- 
matically steering a torpedo by means of a gyroscope which is so 
mounted in the torpedo that the torque introduced by any force 
that tends to turn the axis of the gyroscope causes a precessional 
motion about an axis at right angles to it, and this latter is taken 
advantage of to operate the steering gear, which in turn operates 
the rudder. 

3° Steadying Vessels at Sea. — In the device of Herr Otto 
Schlick for this purpose a heavy flywheel is mounted on a vertical 
axis, which is carried by a frame, which turns on a horizontal 
axle extending athwartship (see Fig. 132). 



218 



DYNAMICS OF MACHINERY 




Fig. 132. 



When the ship starts to roll, a torque is produced about a 
longitudinal horizontal axis, which, instead of causing the axle 

carrying the frame to turn, pro- 
duces a precessional motion of 
the frame about its axle, and 
thus (a) the tendency of the ship 
to roll, and hence the rolling 
force of the waves, is resisted; 
(6) the period of oscillation is 
lengthened and no longer syn- 
chronizes with that of the waves. 
In some experiments made in 
1906, when the rolling, shortly 
before the gyroscope was brought 
into action, was 15 degrees to each side, or a total of 30 degrees, 
it became, when the gyroscope was set free, only 1 degree. In 
order to adapt the apparatus for use in practice, a brake or clamp- 
ing device is added. 

4° Brennari's Monorail Car. — Mr. Brennan employs two gyro- 
stats, both rotating in a vacuum; but the function of the second 
one is to enable the car to go around curves without being pre- 
vented by the gyroscopic action of the first. 

Consider the action of the first. The disk, which, when the 
car is not tilted, is in a vertical fore-and-aft plane, revolves in the 
same direction as the wheels of the car. The axle of the disk is 
horizontal, and is mounted in a frame, which is free to turn about 
a vertical axis, whose bearings 
are attached to the car. On the 
prolongation of the axle of the 
disk are two rollers, at different 
distances from the disk. At- 
tached to the car body are four 
circular guides, suitably placed 
with reference to the rollers. 
When the car is tilted, as by the 
wind, on one side, one of these 
guides presses down on one of 
the rollers, and this causes the 
gyrostat to precess, and this 
precession is accelerated by the 
friction between the roller which 
rolls on the guide and the axle 
of the disk. This causes the car 
to tip back, but as the momentum 
of the car carries it beyond the 
proper position another guide is 
brought into contact with and presses on the other roller, and 
thus the car is caused to return. 



-- 


5 


i 


1 


i 


T 












> 




' 




>' 




Fig. 133. 



BODIES WITH A HIGH ROTATIVE SPEED 219 

There are four of these guides, one to press up, and one down, 
on each roller, and thus the car is finally brought to the proper 
position. 

5° Gyroscopic Compass. — A gyrostat, with its axis horizontal, 
is carried in a frame which floats in a mercury bath. Suppose this 
frame to be placed at any given point on the surface of the earth 
except at the poles. Suppose, also, that the axis of the gyrostat, 
at a given instant, is pointing in some direction other than north 
and south. To fix the ideas, assume it to be pointing east and 
west.* Now, after the lapse of a certain time (however short), the 
effect of the rotation of the earth will be that, were the axis of the 
gyrostat to remain in its original direction in space, it would no 
longer be level, but would be in a tilted position vertically. When 
this occurs (no matter how short the interval), the force of gravity 
develops a torque, and this torque, tending to bring the axis to 
a horizontal position, causes the gyrostat to precess. Moreover, 
this action alwaj^s takes place unless and until the axis of the 
gyrostat points north and south. 

The details of construction will not be explained here, except 
to say that the addition of a clamping device is necessary to 
prevent violent oscillations. 



CRITICAL SPEED. 

Both observation and experiment have repeatedly shown that 
every shaft, when driven at a certain speed lying between a certain 
inferior and a certain superior limit (dependent upon its dimen- 
sions, the modulus of elasticity, and the density of the material 
of which it is made, the loads upon it, and the nature and location 
of its bearings), revolves in a bent form, and, unless suitable 
means are adopted to limit its deflection, the latter might increase 
until fracture ensues. Under these conditions the shaft is said 
to " whirl," and these particular speeds are called " critical " 
speeds. When, however, the speed exceeds the superior limit 
the shaft runs nearly true. 

The loads upon the shafts may consist of disks, pulleys, gears, 
of a steam turbine, or an armature, etc., or the shaft may be 
unloaded, in which latter case the inferior and the superior limit 
coincide. In order to impart a speed greater than the superior 
limit referred to above, it is customary to provide a stop (usually 
in the form of a ring at or near the middle of its length) , to prevent 
the deflection from increasing unduly, while the shaft is being 
accelerated, or else to impart the acceleration at a very rapid 
rate. It is, however, safer to provide a stop even when the second 
method is adopted. 

Two of the analyses given to show why the shaft should 
straighten when the speed exceeds the superior limit will be 

* See Fig. 133. 



220 DYNAMICS OF MACHINERY 

briefly explained here, viz., that of Professor Foppl and that of 
Professor Stodola. 

They both assume, in their discussion of the subject, a disk, or 
other rotating body, fastened to the shaft, and rotating with it, 
whose center of gravity is slightly out of coincidence with the axis 
of the shaft, inasmuch as, at high speeds, a slight eccentricity would 
have much more effect than at low speeds. 

Both discussions are based upon the idea that, inasmuch as the 
driving moment is a couple, and as a couple tends to cause the 
body upon which it acts to rotate around an axis passing through 
the center of gravity of the body, the shaft, when the speed is 
sufficiently high, instead of rotating about the common axis of 
the boxes in which it is held, starts to rotate about a parallel axis 
passing through the center of gravity. 

The following illustration will serve to furnish a brief account of 
the explanation given by Professor Foppl. 

Assume a shaft, running in two bearings, and carrying, at the 
middle of its length, a disk, the center of gravity of the combination 
not coinciding with the axis of the shaft. 

Assume a plane of projection, perpendicular to the shaft at the 
middle of its length. 

Let 0, Fig. 134, be the point where the common axis of the two 

boxes in which the shaft is held 
cuts the plane of projection. 

Let S be the center of gravity 
of the combination, SO being very 
small. 

Let A be the center of the shaft 
at the middle of its length. 

Since the driving moment, as- 
sumed very large, is a couple, the 
combination starts to rotate 
around S. 

S and are, therefore, fixed 
points, and A revolves around S 
in a circle whose radius is SO. 
Hence the deflection of the shaft 
is OA, and its stiffness develops a 
force which acts on the combination, and hence on S, in the 
direction AO. 

Whatever the position of A on the circle shown in the figure, 
the force acting in the direction AO has a component (which 
vanishes when A coincides with 0) in the direction SL. Hence S 
approaches 0. 

On the other hand, the following illustration will serve as a brief 
account of the explanation given by Professor Stodola. When the 
speed is less than the inferior limit, centrifugal force causes the 
deflection OA (Fig. 135) to increase, this in turn causing the cen- 




BODIES WITH A HIGH ROTATIVE SPEED 221 

trifugal force to increase, and so on. Then, neglecting the weight 
of the shaft, let 

a = rotative speed in radians per second. 
W = weight of disk in pounds. 

W 

m = — = mass of disk. 

9 

OA = y = deflection. 

AS = e = eccentricity. 




Then we obtain, in accordance with the theory of beams, 

48 EI 



ma 2 (y + e) = ay, where a in this case = 



••• y= XT7' (1) 

ma 2 

Hence, as a increases, and fracture would eventually result were 
not the deflection prevented from becoming excessive, the means 
most commonly employed being a fixed ring at or near the middle 
of the length. When this is done, and when the speed increases 
until it has become greater than the superior limit referred to 
above, a new condition of equilibrium occurs, at which the point 
A, as it were, exchanges places with the center of gravity S, inas- 
much as the shaft starts to rotate about an axis through S. We 
thus have 

ma 2 (y — e) — ay; 



V = 



a 



mo? 



Then we find that as a increases y decreases. 

Moreover, analogies are to be found, in the whirling of a shaft, 
with the case of the inflexional elastica, i.e., with that of a column 
of so great a length that the stable form is a curvilinear one. 
Other analogies will be found in the case of a rod vibrating trans- 
versely. 

In determining by calculation the critical speed, or speeds, of a 
shaft of given dimensions, and carrying given pulleys, etc., we 
merely consider the shaft, at any one instant, as a beam, having 
for loads the various centrifugal forces to which it is subjected, 
and find the speed of rotation required to secure equilibrium under 
the conditions of support, etc., existing. 



222 DYNAMICS OF MACHINERY 

Moreover, the results of experiment bear out very closely those 
of such computations. 

Formulae for the critical velocity of unloaded shafts supported 
in different ways were deduced many years ago by Professor 
Rankine and by Professor Green hill. 

In 1894 an article appeared in the Transactions of the Royal 
Philosophical Society of London by Mr. Stanley Dunkerley, pre- 
sented by Prof. Osborne Reynolds, in which are contained: 

(a) An extension of the theory by Professor Reynolds to the 
cases of shafts loaded with one or more pulleys; 

(b) An account of a series of experiments of this kind made by 
Mr. Dunkerley; 

(c) A comparison of the experimental results with those ob- 
tained by calculation. 

Moreover, the greatest discrepancy found in any one case was 
8.7 per cent of the observed value, and in almost all cases it was 
far less. 

In the Civil Ingenieur for 1895 there are two articles by Pro- 
fessor Foppl on the subject of critical velocity, and, in the same 
volume, there is an article by Mr. L. Klein, giving an account of 
a series of experiments made by him upon the critical velocity of 
shafting. 

FORMULAE. 

Let A = area of cross section of shaft in square inches. 

J = moment of inertia (units being inches) of cross section 

about a diameter. 
a = critical speed in radians per second. 
w — weight in pounds of one cubic inch of the material. 
g = 386 inches per second = acceleration due to gravity. 
E = modulus of elasticity of the material in pounds per 

square inch. 
W = weight in pounds of any pulley which the shaft carries. 
I = length of shaft assumed supported in bearings at its 
ends. 

Then we shall have : 

1° For an unloaded shaft of length I, merely supported in its 
two end bearings: ^2 njjjjj 

a = P\^A (1) 

2° For a shaft carrying a disk or pulley at the middle of its 
length, and merely supported in its two end bearings, if the weight 
of and the centrifugal force of the shaft itself be neglected : 

a = V~fFP~ (2) 

A deduction of 1»hese formulae, together with a brief account of 
the method pursued by Professor Reynolds, will be found in 
Appendix C. 



APPENDIX A. 

Principal Axes and Moments of Inertia. 

Assume the body to be referred to three rectangular axes OX, 
OY, and OZ (Fig. 23), and let it be required to find the principal 
axes of inertia and the moments of inertia about these axes. 

Use the second definition, i.e., that the principal moments of 
inertia must satisfy the conditions to render them maximum or 
minimum. As has been shown, we have, for the moment of 
inertia / about the axis OV, which makes angles, a, ft, and 7 re- 
spectively with OX, OY, and OZ, 

I = A cos 2 a + B cos 2 ft + C cos 2 y — 2D cos ft cos y 

— 2 E cos a cos 7 — 2 F cos a cos ft. . (9) 

The angles a, ft, and y, are, however, subject to the condition 

cos 2 a + cos 2 ft + cos 2 7 = 1 (10) 

But, since J must be a maximum or minimum, dl = 0, hence* 
differentiating (9) and reducing, we have 

(A cos a — F cos ft — E COS7) sin a da -f- ( — ^ cos a-\- B cos 
— D cos 7) sin ft dft + ( — £7 cos a — D cos 

+ Ccos7)sin7^7 = 0. . . . (11) 

By differentiating (10) we obtain 

(cos a) sin a da + (cos ft) sin ft dft -f (cos 7) sin 7 cfy = 0. (12) 

Multiply (12) by an undetermined multiplier X, and subtract the 
result from (11). We thus obtain 

{ (A cos a — F cos (3 — E cos 7) — X cos a\ sin a da 
+ J( — ^ cos a + i? cos ft — DCOS7) — XcosftS smft dft 
+ {(— l^cosa — D cos ft + C COS7)— X cos 7} sin7C?7 = 0. (13) 

Then equating to zero the coefficients of da, dft, and dy respectively 
we have 

_ A cos a — F cos ft — E cos 7 — Fcosa +#cosft — Dcos7 
cos a cos ft 

— E cos a — D cos ft -f- C cos 7 ,.. , x 

= (14) 

cos 7 

223 



224 APPENDIX A 

From these we obtain 

_ A cos 2 a — F cos a cos £ — E cos a cos 7 
cos 2 a 

— F cos a cos |8 + 7> cos 2 /3 — 7) cos /3 cos 7 

cos 2 jS 

— Z£ cos a cos 7 — 7) cos |8 cos 7 + C cos 2 7 /ir\ 

cos 2 7 

Hence, adding numerators for a new numerator, and denominators 
for a new denominator, we have 

A cos 2 a+B cos 2 /8+C cos 2 y-2 D cos /3 cost-2 i? cos a cosy-2 F cos a cos ^ 

cos 2 a+ cos 2 /3 + cos 2 7 

Hence, substituting 7 for X in (14) and reducing, we obtain the 
three equations: 

(A - 7) cos a - F cos /3 - E cos 7 = 0, . . . (16) 

- F cos a + (5 - 7) cos j8 - 7) cos 7 = 0,. (17) 

- # cos a - 7) cos j8 + (C - 7) cos 7 = 0,. (18) 

which, together with cos 2 a + cos 2 j8 + cos 2 7 = 1, furnish the 
solution of the problem. 

The solution can be effected as follows : Eliminating cos a, cos /5, 
and cos 7 from (16), (17), and (18), we obtain 

(A -I)(B- 7) (C - I) - 7) 2 (A -I)- E* (B - I) 

- F 2 (C - 7) - 2 DEF = 0, (19) 

which may be written 

P - (A + B + O 7 2 + {BC + AC + AB-D 2 - E 2 - F 2 ) I 

- {ABC - D 2 A - E 2 B - F 2 C - 2 DEF) = 0. (20) 

Solving this cubic, we obtain the values of the three principal 
moments of inertia. 

Hence there are three principal moments and three principal 
axes of inertia. Having thus found the values of the three prin- 
cipal moments of inertia, which will be called 7i, 7 2 , and 7 3 respec- 
tively, we next proceed to find the angles made by the corre- 
sponding axes, «i, /Si, 71, being those made by the axis about which 
7i is the moment of inertia, with OX, OY, and OZ; a 2 , ^2, and 
72 corresponding to 7 2 , and az, /3 3 , 73 corresponding to 7 3 . By 
substituting 7i for 7 in equations (16), (17), and (18) and solving, 
we obtain 

cos ai _ E (B - h) + FD F (C - 7i) + DE 

. (21) 



cos 71 (A -h)(B -70 -F 2 D(A-h) + EF 
(B- 7Q (C - Id - D 2 

E(B-h)+FD ' 



APPENDIX A 225 

cos ft ^ D (A - h) + EF (A - h) (C - h) - E 2 

cos 71 (A - 7i) (B - h) -F 2 D{A- h) + EF 

_ F(C-h)+ED 

~ EiB-IO+FD' {ZZ) 

which equations, combined with 

cos 2 on + cos 2 0i + cos 2 Yi = 1, . . . . (23) 

give us the values of ai, 0i, and 71 respectively. 

The values of a 2 , 02, and 72, and of 0-3, 03, and 73, can be found 
in like manner, by substituting in (21), (22), and (23), for the 
subscript 1, the subscripts 2 and 3 respectively. 

Solution of Examples. 

In many practical applications, we already know one of the 
principal axes, and the corresponding moment of inertia, and then 
the solution is more easily made by the method already outlined 
on pages 22, 23 and 24. 

To illustrate a case of this kind, Example I, page 31, will be 
worked out by the general method. 

The following values were calculated when the example was 
solved by the former method : 

A = 150.24, B = 158.24, C = 55.52, D = F = 0, E = 4.00. 

From these we obtain 

A+B + C = 364.00 

BC + AC + AB - D 2 - E 2 - F 2 = . . 40,884.79 
ABC - D 2 A - E 2 B - F 2 C - 2 DEF = . 1,317,399.53 

Hence the cubic becomes 

P - 364 P + 40,884.79 I - 1,317,399.53 = 0. 

One root of this is I = 158.24. .*. Dividing out by I- 158.24 
we have 

P- 205.76 I = - 8325.33. 



/. I = 102.88 ± V2258.96 = 102.88 ± 47.5285. 
/. 7i= 150.4085, 1 2 = 158.2400, 7 3 = 55.3515. 

To calculate the angles a h 0i, 71; a 2 , 02, 72; and a 3 , 03, 73, pro- 
ceed as follows: 

7i = 150.4085, A - h = 0.1685, B-I 1 = 7.8315, 
C - 7i = - 94.8885. 

Hence from (21) and (22) 

cos ai _ — 4 1 cos a\ _ 

cosTi ~~ 0.1685 " = ~ 0.0421' ° r cos 71 ~ o' 



226 APPENDIX A 

cos «i 94.8885 1 



or 



cos 7i 4 0.0421 

cos 0i _ cos 0i cos /Si _ 
= 0, or = -j or = 0. 

COS Yi COS yi COS Yl 

.*. cos 0i = 0. .". sin a\ = cos yi- •'• tan an = — 0.0421. 

«i=.2°25', /3i = 90°, 71 = 92° 25'. 

7 2 = 158.24, A-7 2 =-8.00, B-I 2 = 0.00, C - I 2 = -102.72. 

cos a 2 cos a 2 cos o; 2 
= S ! or = -> or = 7;- 

COS Y2 COS Y2 COS Y2 

cos 02 cos 02 cos 02 
= -> or — - — = 00, or = -• 

COS Y2 COS Y2 COS Y2 

.'. cos Y2 = 0. /. cos a 2 = 0. /. a 2 = 90°, 2 = 0°, Y2 = 90°. 
7 3 = 55.3515, A-J 3 = 94.8885, B-I 3 = 102.8885, C-J 3 = 0.1685. 

COS az 4 „ A . 01 COS Q! 3 

= c\a ootr = 0.0421, Or = 7:' 

cos Y3 94.885 cos 73 

cos«3 = 01685 = Q Q421 
cos 73 4 

COS 03 n COS 3 COS 03 A 

= 0, or = -1 or = 0. 

cos 73 cos 73 cos 73 

.*. cos 03 = 0. .*. cos 0:3 = sin 73. .*. tan 73 = 0.0421. 

a 3 = 87°35', 03 = 90°, 73 = 2°25 / . 

This example has been solved by this method, in order to illus- 
trate the application of the general method to a special problem. 
The method itself is, however, unnecessarily long in a case of this 
kind, when one of the principal axes coincides with one of the 
original axes OX, OY, or OZ. 

Moreover, the fact that the moments of inertia differ but little 
from the moments of inertia about the original axes renders it 
impossible to obtain a good degree of accuracy in the angles, 
without making use of an unduly larger number of 1 decimal places 
than is warranted by the data. The example in question can be 
solved much more satisfactorily by the method outlined on 
pages 22, 23 and 24. 

Perpendicularity of the Principal Axes. 

Equations (16), (17), and (18) may be written 

F cos + E cos 7 = (A — I) cos a. . . . (24) 

F cos a + D cos 7 = (B - I) cos 0. . . . (25) 

E cos a + D cos = (C - I) cos 7- • • • (26) 

EF . DF 

To both sides of (24) add -yr- cos a, to both sides of (25) -=- cos j3 r 

jj hi 



APPENDIX A 227 

DE 
and to both sides of (26) —=r cos 7, and collecting the terms and 

r 

factoring, we have 

EF (^ + ^ + c ^p) = [D (.4 - I) + EF] «»«. (27) 

Z)f(^ + ^ + ^) = [S(B-/)+OF]^. (28) 

DB(^ + °°^ + !°pL)-[F(C-Z)+DB\2p. (29) 
From these we readily deduce 



cos a 



EF D 



D (A — I) + EF cos a cos ft cos 7 

cos ft 
DF E 



E {B — I) + DF cos a cos ft cos 7 

cos 7 



F (C — I) + DE cos « cos ft cos 7 

D + E + F 

Also the following: 

1 cos a 



D{A — I) + EF cos a cos ft cos 7 

D + # + F 

1 cos ft 



E {B — I) + DF cos a cos ft cos 7 
1 cos 7 



F (C — I) + D# cos os cos ft cos 7 

D + E +_ F~ 



(30) 



(31) 



(32) 



(33) 



(34) 



(35) 



Adding (30), (31), and (32), we have 

EF , DF DE 



Z) (A - 7) + EF ' E(B-I) + DF ' F (C - I) + DE 



228 APPENDIX A 

Substituting, in (36), respectively 1 \ and Is in place of /, and 
subtracting, and then dividing out by DEF (Is — I 2 ) we obtain 

1 1 1 1 

D(A-I 2 )+EF' D(A-h) + EF + E(B-I 2 )+DF' E(B-h)+DF 

+ F(C-I 2 )+DE ' F(C-h)+DE = ° (37) 

Substituting in (37) the values obtained from (33), (34), and (35), 
we obtain 

COS a 2 COS Q!3 + cos 2 cos 03 + cos 72 cos 73 = 0, . . (38) 

which expresses the fact that the cosine of the angle between the 
axes of I 2 and Is is zero, hence that these axes are at right angles 
to each other. 

In the same way it may be shown that 

cos ai cos «3 + cos 0i cos 03 + cos 71 cos 73 = 0. . . (39) 

COS ai COS (22 + cos 0i cos /3 2 + cos 71 cos 72 = 0. . . (40) 

Hence the three principal axes form a rectangular system. 

Products of Inertia about Principal Axes. 

Let x, y, z be the coordinates of any point in the body when 
referred to OX, OY, and OZ (Fig. 23), and x h y h z h those of the 
same point when referred to the principal axes of inertia. Then 
we have 

Xi=x cos a\-\-y cos 0i+2 cos 71, yi= x cos a 2 -\-y cos 02+2 cos 72 
and Z\= x cos 0:3 +2/ cos 03 +2 cos 73. 

Then we have 

V\Z\ = yz (cos 02 cos 73+ cos 03 cos 72) -\-xz (cos a 2 cos 73+ cos az cos 72) 
-\-xy (cos a 2 cos 03+ cos a 3 cos 2 ) +£ 2 cos a 2 cos 0:3 
-\-y 2 COS 02 COS 03+2 2 cos 72 cos 73, 

and since 

cos a 2 cos a s = — (cos 02 COS 03 + cos 72 cos 73) , 
COS 02 cos 03 = — (cos a 2 cos as + cos 72 cos 73), 

and cos 72 cos 73 = — (cos a 2 cos as + cos 2 cos 3 ), 

we obtain 

ViZi = yz (cos 02 cos 73+ cos 3 cos 72) +xz (cosa 2 cos 73+ cos as cos 72) 
+xy (cos a 2 cos 3 +cos 0:3 cos 2 ) 

— x 2 (cos 02 cos 03+cos 72 cos ys)—y 2 (cos a 2 cos a 3 +cos 72 cos 73) 

— Z 2 (COS a 2 COS as + COS 02 COS 03) 

= yz (COS0 2 COS 73 + COS03 cos 72) +xz (cos a 2 cos 73 + cos as cos 72) 
-\-xy (cos a 2 cos 3 +cos 0:3 cos 2 ) 

— (y 2 +z 2 ) cos a 2 cos as — (x 2 +Z 2 ) COS 2 COS 03 

— (x 2 +y 2 ) cos 72 cos 73. 



APPENDIX A 



229 



Hence since D\ — limit of 'LwyiZi, we have 

Di =D(cos 2 cos 73 + cos 3 cos 72) + 2?(cos a 2 cos 73+ cos 0:3 cos 72) 

+ F(cos a 2 COS 3 + COS a 3 COS 2 ) 

—A cos a 2 cos as — B cos 2 cos 03 — C cos 72 cos 73. 

Corresponding results obtain for E\ and i'Y 

Each of these may be expressed in two ways, as follows: 

Di = — (A cos a 2 — F cos (3 2 — E cos 72) cos 0:3 

— ( — F cos a 2 -\- B cos0 2 — 2) cos 72) cos 3 

— (— E cos a 2 — D cos 2 + C cos 72) cos 73. 






Di = —(A cos a 3 



&=- 



Ei=- 



F, = - 



Fi = - 



2^ cos 3 — 22 cos 73) cos a 2 

""■ 03 — 2) cos 73) cos j8 2 



— F COS a'3 + B cos /3 3 — 2) cos 73) COS 02 
^— E cos az — D COS 03 + C cos 73) cos 72. 
A COS ai — F COS 0i — 2? COS 71) COS 0:3 
[— Fcosai + B cos 0i — 2) cos 71) cos 03 
'— E cos ai — D cos 0i + C cos 71) cos 73. 

A cos as — F cos 03 — 2? cos 73) cos ai 
jP COS 0:3 + B COS 03 — D COS 73) COS 0i 

E cos 0:3 — D cos 03 + C cos 73) cos 71. 



— E cos 0:3 
A cos ai — F cos 0i — E cos 71) cos a 2 

— F cos ai + B cos 0i — 2) cos 71) cos 2 

— E cos ai — D cos 0i + C cos 71) cos 72 . 
A cos a 2 — F cos 02 — 22 cos 72) cos ai 

— F cos a 2 -\- B cos 02 — 2) cos 72) cos 0i 

— 22 cos a 2 — D cos 02 + C cos 72) cos 71. 



uos a 2 — u uus P2 ~r ^ eus 72; eus 71. 
Moreover, since the three principal axes form a rectangular system, 

cos a 2 cos a: 3 + cos 02 cos 03 + cos 72 cos 73 = 0, 
cos ai COS 0:3 + COS 0i COS 03 + cos 71 cos 73 = 0, 
COS «i COS a 2 + cos 0i COS 02 + COS 7i i 

From equation (14) we have 



cos 72 = 0. 



A cos a\ — F cos 0i — E cos 71 = Xi cos a h 

— F cos a\ + B cos 0i — 2) cos 71 = Xi cos 0i, 

— E cos ai — D cos 0i + C cos 71 = Xi cos 71. 
A cos a 2 — F cos 02 — E cos 72 = X 2 cos a 2 , 

— F cos a 2 -\- B cos 02 — 2) cos 72 = X2 cos 2 , 



A COS (23 



— F cos a 2 -\- B cos 2 — 2) cos 72 = X2 cos 2 , 

— E cos a 2 — D cos 02 + C cos 72 = X 2 cos 72. 
- F cos 03 — E cos 73 = X3 cos 0:3, 

F COS a 3 + B COS 03 — D'COS 73 = X3 COS 03, 
— E COS (X"i — ^ r»nc /3„ —I— 1! nr\a »«i/„ := \« nr»e <>/„ 



F COS a 3 + 2? COS 03 — D'COS 73 = X3 COS 03, 
- E COS as — D COS 03 + C COS 73 = X3 cos 73. 

Substituting these values for the parentheses in the values of 
Dij E h and F h we have 

cos 72 cos 73) = 0, 



£L 1, ana r 1, we nave 

2)i = X2X3 (cos a 2 cos 0:3 + cos 2 cos 3 + cos 72 cos 73) = 0, 
Ei = X1X3 (cos ai COS 0:3 + COS 0i COS 03 + cos 71 cos 73) = 0, 
Fi = X1X2 (cos ai cos a 2 + COS 0i cos 2 + cos 71 cos 72) = 0. 



230 APPENDIX A 

Hence it follows that when the principal axes of inertia are taken 
as coordinate axes, the products of inertia are zero. Therefore, 
the two definitions given per principal axes, and for principal 
moments of inertia, are equivalent. 

Momental Ellipsoid. 

Equation (1) gives, for the moment of inertia / of a body 
about an axis which makes, with OX, OY, and OZ, angles a, /3, and 
7 respectively, 

I=A cos 2 a-\-B cos 2 (3+C cos 2 7 — 2 D cos /3 cos 7 — 2 E cos a cos y 

— 2Fcosacosp.. . (41) 

A geometrical interpretation of this equation can be obtained as 
follows: find the equation of a surface, the square of whose 
radius vector p in any given direction, making angles a, /3, y 
respectively, with OX, OY, and OZ, shall be equal to the quotient 
obtained by dividing a positive arbitrary constant p by the 
moment of inertia of the body about this radius vector taken as 
axis. We thus have, if x, y, z be the running rectangular co- 
ordinates of the surface, the relations 

x — p cos a, y = p cos (3, z = p cos y, p 2 = x 2 + y 2 + z 2 = j 



T 



Ip 2 = p., and /■=-=■ 



To deduce the equation of this surface, multiply each side of 
equation (41) by p 2 and we shall have 

7p 2 = Ax 2 + By 2 + Cz 2 -2 Dyz - 2 Exz - 2 Fxy. 

And since Ip 2 = p., the equation of the surface becomes 

Ax 2 + By 2 + Cz 2 -2 Dyz - 2 Exz - 2 Fxy = p. . . (42) 

This surface is, moreover, one which may be said to represent the 
moments of inertia of the body about all axes through the origin; 
for, if we wish the moment of inertia of the body about an axis 
through the origin, which makes, with OX, OY, and OZ, the angles 
a, (3, and y, we have only to determine the radius vector of the 
surface having this direction, square it, and divide p by p 2 , and 
we shall have the moment of inertia desired. Equation (42) is the 
equation of a quadric, and, since the moments of inertia of a body 
about all axes are positive, it follows that p 2 is always positive, 
hence that p is never imaginary. Hence it follows that the sur- 
face is an ellipsoid. It is called the Momental Ellipsoid. 

It follows, therefore, that there are three principal axes at 
right angles to each other, and that the moments of inertia about 
these axes fulfill the conditions necessary to render them maxi- 
mum or minimum. Indeed, all the previous propositions could 
have been deduced from the properties of the momental ellipsoid. 






APPENDIX A 231 

When OX, OY, and OZ are the principal axes, then A, B, and C are 
the principal moments of inertia, and D = E = F = 0. In that 
case, the equation of the momental ellipsoid becomes 

Ax 2 + By 2 + Cz 2 = fi (43) 

Ellipsoid of Gyration. 

~ The reciprocal surface to the momental ellipsoid is called the 
Ellipsoid of Gyration; thus, when the axes OX, OY, OZ are prin- 
cipal axes, its equation is 

/>»2 *iy2 2j2 

-r + %z + -pz = sl constant. 
ABC 

Many interesting propositions about moments of inertia can be 
deduced from a study of these two ellipsoids, but the subject will 
not be pursued further here. 



APPENDIX B. 

Deduction of the Formulae for Governor Oscillations. 
1° Reciprocating Parts of Valve Gears. 

The motion will be considered harmonic. 

Resistance = Pq sin (at + 8). 

Component of resistance along relative path of eccentric 

= Pq sin (at + 8) sin at. 

.'. Moment of resistance = M — P ri sin (at + 8) sin at. 
This reduces to 

M =^p{cos5 - cos (2 a* + 8)\. 

2 IT Pari { 1 ) at = 2 ir 

— M m = Ti — ] <*t cos 5 — o sin (2 at + 6) > 

a 2a ( J )a* = o 

= —p— (2 7r cos 5). .. M m = — s- cos 5- 
/. Mi = M - M m = -^pcos (2af + 5). 



= r^y^r 5 sin 8 - sin (2 at + 8)\. 



4W P 2 a 

2ir Q P gri i . . . , 1 , Q ,, ^? a ' = 2,r 

— m = -—=■ — — alsmHo cos (2 at + 5) > 
a 4 14 p^a z ( Z ) a t = o 

= TjW (2 " sm6) ' 

= -^^Psin 8. .'. ^ = - m = - ^^ sin (2 a* + 5). 
4lW p 2 a 4 M/p 2 a 



-I 



' Pn /7ri J atf= aJ 

^1^ = ^-^008(2^+5) 

8 It p^a 2 )a<=o 

- snk-?l S - cos 8 + cos (2a< + 5)} 
232 



APPENDIX B 233 



2lT Pq gri ( 1 )a* = 2ir 

— rim = ^™ -5-5 } — <*t cos 5 + 5 sin (2 a* + 5) 

a O KK P^a J f Z 



. a< = 

A grri / ^ 

8^7^( 27rCOs6 )- 



p"a' 

When 0i = 0, sin (2a* + 5) = 0. .*. cos (2a* + ,5) = 1 or - 1. 

Hence maximum positive value of yji = 7^7-^-7 • 

8 W n2 ™ 2 



p"a 
Hence maximum negative value of tji = 



Po gri 



SWp 2 a 2 

Hence greatest travel = -t-^Tt "irr 

4 W p L cr 



2° Steam Pressure on End of Valve Rod. 

2 7r 1 ) 27r 

M = —P 2 r i sin at. .'. — M m = - P 2 r x cos at > =0. .*. M m = 0. 

« « ) o 

.*. Mi = M - M m = - P 2 r 2 sin at. 



= f rMi^^-f-'cosa^" '^-^(-l + cosa*). 
/Jo W p 2 a )at = o Wp 2 a x 

— 0m = w -t-o ( — od + sin a*) > 
a *r p z a z )a< = o 

a \W p 2 a) ' W p 2 a 



PiQr^ 



e x == e — e m = =tS 9 cos at - 

W p 2 a 



-I 



75 07*1 ) at= t 

Sidt= — ^f \-z sin at > 

W p 2 a 2 )at = o 

Pi gri . , 
sin at. 



W P 



2 a 2 



27T PiflTi W-2* 

a JFp 2 a 3 \ at » 



P2 gri . , 
.. Y]i = Y) — r\m= — ^ -7-5 sin a*. 

W p 2 a 2 

When 6i = 0, at = » or a£ = -=- • 

.*. Greatest positive value of tqi = ^| -t-^ * 

H/ p l a l 



234 APPENDIX B 

Greatest negative value of m = — tJ-^-A • 

W p'a' 

.'. Greatest travel = 2 ^ -^-A, • 



3° Weight of Valve, Valve Rod, and Eccentric Rod. 

The deductions in this case are identical with those of No. 2, 
except that P% is substituted for P 2 throughout. 

4° Friction of Valve and Valve Rod. 
at < jr — 5; ikf = — Piri sin atf. 

TV . 

at > -z — 8; M = Ppi sin at. 

ir 1 ) a ' = i~ 5 1 i<*t=ic 

- M m = - PiTi cos atf > = Ptfi cos atf > 

2 Pin sin 5 ,, 2Pirisin5 

= • .*. M m = • 

a tv 

at<^- 8; M 1 = M-M m = P 1 r 1 (- sin a* - ^1\ . 

aZ > - - 5; M x = M - M m = P x r x f sin a* J ■ 

«*<» 5 . e = g j r Ml dt = ^\-l + cosat-2at^l. 
2 I Jo W p 2 a ( tv ) 

*>I-J; ,-§£->(- 1+2*"**) 

2 Tr p 2 a\ 7r / 

, ^i^i ( , c^sinS^ 

+ w V } ~ cos arf - 2 > 

KK p 2 a ( tv ) 

Pi gr\ ( . . _ . _ , . sin 5 ) 
.-. ^ = — i *— J — 1 + 2 sin 5 — cos at — 2 at > 

IT p Z « ( TV ) 

« W p 2 a 2 \ tv / at = 

+ ( — at + 2 at sin 5 — sin atf — a 2 £ 2 ) 



««- s -« 



■jr 



r a* = 



«<=--* 



•"• - e " = W-Tli 2 (cos5 + 5sin5) - tt 
a It p z a z / 



APPENDIX B 235 

. Pi gri { 2 (cos 5 + 5 sin 5) ) 

" Vm ~W P 'a( it M ' 

z^ 71 " x- ft - ft ft - Pl 9 Tl S 9 cos_5_-Msin5 
2 It p 2 a( ir 

sm8) 
+ cos at — 2 at > • 

7T ) 

7r . „ n n Pi gri ( n . „ n cos 8 + 5 sin 5 

a<>S-5; 0i = 0-0™ = ^ V-] 2sm5 - 2 — 

2 It pa / x 



— cos at — 2 at 



7T ) 

cos 5 + 8 sin 5 



at<l-8;r}= ( l eidt = ^^\\-2at 
2 Jo 1^ pa f 



7T 



+ sin at — aH 2 

7T 



,^tt _ Pi^riCf , .cos5 + 5sin5 

at> 2- 8;r] = W?a-*)r {T ~ 28) ~ 



+ cos 5 



-(H 



TT 

sin5~| 



, T_ J . . _ . cos 5 + 5 sin 8 
+ 2 at sin 5 — 2 atf sm ctf 

-at = t x 






i" 1 



.*. f) = ^ ^ < 2 (cos 5+5 sin 5) —w sin 6+2 at sin 5 
It p l a l ( 

» , cos 5+5 sin 5 . 0J0 sin 6 J 
-2a^ sin at — aH 2 > • 

IT IT ) 

a W p z a 6 ( |_ 7r 



.a«=^-« 



_a^sin5"| +r2^(cos5 + 5sin5) 

O ^ Ja* = L 

. • s , 2 ,2 • s 2 ., cos 5 + 5 sin 5 
— 7ra2 sin 5 + aH 2 sm 5 — a 2 £ 2 \- cos a£ 

7T 



a 3 2 3 sin 



E5il I 

* Jrf-f-*} 

-*» = W ■?! i (- 2 - T5 + 62 ) sin 5 2 +5 cos 5 I • 
a W p 2 a 3 f \ 12 / \ 






236 



APPENDIX B 



T]to = 



l Pi gn 



7T 



2 - — + 5 2 lsinS + 2 5cos5 



7T W p 2 a 2 

Then for the value of yji in each case calculate tq and subtract 
rim since yji = r t — Y] m . 

How to find the greatest travel has already been explained. 



5° Action of Gravity on the Swinging Weight. 

From Fig. 136, we readily see that the perpendicular distance 
from P to a vertical line through the center of gravity of the 
swinging weight is x x sin (at — ($), where Xi = distance of center 
of gravity of swinging weight from pivot. 




Fig. 136. 

M = W Xi sin (at - &) . /. — M m = ^^ cos (at -0)" = * 

a a at = o 



M m =0. 
/ 



Mi = M - M m = Wxi sin (at - (3). 



g jpMidt = 



gxi 



p*a 



_ R) ^=at == gx ls 

at = p 2 a 



cos (at- 0) ' '=*£Mcosj8-cos(<rf-j8)} 



— e m = g p- 2 \at cos j8 - sin (at - j3)} 
a o l a l 



at = 2ir 
a* = 



= ^(27rcos«. .'. ^=^008/3. 



p 2 a 2 



p 2 a 



APPENDIX B 237 



/. 0j = e - 6 m = - 9 -p cos (at - |8). 

p L a 

^ = r^d«= -^ 2 sin(^-/3) 
Jo p^ 



at = 



= —^Y fsin/3 + sin (at - 0) J. 

.. — iQm = -\-JatsmP - cos (a:Z - 0) = V^sm/3. 

a pa «* = o a p 2 a 2 

.*. ?)m = ^4 sin 0. 

Y)i = tq — r} m = — ■ sin (at — )8). 

P^CT 

When 0i = then cos (a* - 0) = 0. 

.*. «$ = - + £ or a< = -y + 0. 
Hence: Greatest positive value of tqi is -^-4 » 

and Greatest negative value of tqi is — -^-^ • 

p J a z 

Greatest travel of swinging weight = — ^ • 

p r cr 



APPENDIX C. 

Critical Velocity of Shafting. 

Let OX be the original alignment of the shaft. 

OF be perpendicular to OX, lie in the plane of bending, and 
revolve with the shaft. 
M = bending moment, in inch-pounds, at distance x from 
0, in inches. 
y = deflection, in inches, at distance x from 0, in inches. 
A = area of cross section of shaft, in square inches. 
J = moment of inertia (units being inches) of cross section 

about a diameter. 
a = angular velocity of shaft, in radians per second. 
w = weight, in pounds, of one cubic inch of the material. 
W a = wAl = total weight of shaft, in pounds. 
g = 386 inches per second = acceleration due to gravity. 

11) A Df 

C = y — centrifugal force, in pounds, of one-inch 

y 

length of shaft, at distance x from 0, 

in inches. 

E = modulus of elasticity of the material, in pounds per 

square inch. 
W= weight, in pounds, of any pulley that the shaft carries. 
7'= moment of inertia (units, pounds and inches) of pulley 

about a diameter in its middle plane. 

From the common theory of beams we obtain the two following 
equations : 

■3F-C-— * • • CD and ^ 2 = m . ... (2) 

By differentiation we obtain 



d*y /Q , , ; fW Ac? 

dtf= m y < 3 ) where m= \~g~ET' 

The general solution of equation (3) may be written 

y = Ae mx + Be~ mx + C cos mx + D sin mx, . . (4) 

A, B, C, and D being constants to be determined from the con- 
ditions of the problem. By differentiation we obtain 

-r- = m \Ae mx — Be"™* — C sin mx + D cos mx\ . . (5) 
dx 

238 



APPENDIX C 239 

t^ = m 2 \Ae mx + Be~ mx — C cos mx — D sin mx\ . . (6) 
ax 2 

In the case of shafts that are unloaded, and are not continuous over 
the bearings, the above equations suffice for the solution. 

When, however, the shaft carries a pulley, or when it is con- 
tinuous, we have to do with concentrated loads, and the following 
conditions hold: 

1° In the case of a pulley 

Let R = bending moment to the right of the pulley, and close to it. 
L = bending moment to the left of the pulley, and close to it. 

Then, when the pulley is on a portion of the shaft that is inclined to 

dv 
OX, the tangent of the angle of inclination being -^- > inasmuch as 

the plane of the pulley is part on one side and part on the other 
of a plane through its center perpendicular to the original axis of 
rotation, the centrifugal forces of the two parts form a couple 
whose moment can readily be shown to be approximately 

FoPdy 

g dx 
Hence we have 

R - L = l^f (7) 

g dx 

Moreover, since -r- and -j- are the shearing forces respectively to 

the right and left of the point where the pulley is attached to the 
shaft, we shall have 

dR dL Wa 2 

Tx~Tx-~T y (8) 

2° When the shaft is continuous, if 

R = bending moment just to the right of support, 
L = bending moment just to the left of support, 
S = supporting force, 

we shall have 

— — — =- S (9) 

dx dx 



Unloaded Shaft, Length I, merely Supported on its Two End 

Bearings. 

Take the origin at the left-hand support. Then, when 

x = 0, and when x — 1, 

d 2 y 
we have y = and -j\ — 0. 



240 APPENDIX C 

Imposing these conditions on equations (4) and (6), we obtain 

A+B + C = (10) 

A + B -C = (11) 

Ae ml + Be-™ 1 + C cos ml + D sin ml = 0. . . (12) 

Ae ml + Be-™ 1 - C cos ml - D sin ml = 0. . . (13) 

e^ ma 



Fig. 137. 

Equations (10) and (11) give 

C = and £ = - A. 

By substitution in (12) and (13) 

A (e ml — e-™ 1 ) + £) sin ml = 0. 
A (e m * — e~ ml ) — D sin ml = 0. 
Hence A (e ml — e-™ 1 ) = and D sin mZ = 0. 

One solution is when A=B = C = D = 0, and hence the shaft 
does not whirl. 

One solution is when e ml = e-™ 1 i.e., when 1=0, and the shaft 
does not exist. 

Hence the only solution that applies is when 

sin ml = (14) 

As neither m nor I is zero, we must have 

ml = 7r, or an odd multiple of ir. 
Moreover, 

ml = ir gives «=7JV^A < 15 > 

N 
If N — number of revolutions per minute, a = ^ ; hence 

"=^v/f. «. 

This may be written 

Art/^i - u z^ 900 7T 2 ^/ 

z yw,' re i = — v — 

Shaft Carrying a Pulley, the Weight and Centrifugal Force of the 

Shaft being Neglected. 

In these cases m becomes zero, and hence equation (3) becomes 

S=° < 17 > 



APPENDIX C 241 



The general solution o. this equation is 



y = ^rx* + ^x*+Cx + D (18) 



By differentiation we obtain 

g_|* + Bs + c (19) 

% = Ax + B (20) 

Moreover, with a non-continuous shaft we have also 

fl _ L = 7V^ 

g ax 
dR dL Wa 2 
dx--dx- = ^~ y < M > 

By this means Professor Reynolds worked out many cases of 
shafts carrying a pulley in different positions. The work is often 
long, but in certain cases it can be very much shortened. 

Example. — Find the critical speed of a shaft, merely supported 
in its two end bearings, and carrying a pulley or disk at the middle 
of its length, neglecting the weight of and the centrifugal force of 
the shaft itself. 

Solution. — Considering the shaft as a beam supported at the 
ends and loaded at the middle of its length with a load P, the 
deflection at the middle would be 

I P/ 3 
In this case, however, we have 



y 48 EI 



D Wa 2 

Therefore 

I Wa 2 l* 
y . 48 gEI V ' 
:. the critical speed is 

/ 48 gEI 

If N = number of revolutions per minute, a = — • 

■•■ «^w <*> 

This may be written 

m J® 2 i, v 48 gEI (900\ 

N =\w> where E2 = -wir{^-)' 



INDEX 



PAGE 

Absorption dynamometers 13 

Acceleration, graphical construction 100 

Accelerations, parallel to line of dead points, — tables 52 

Action of reciprocating parts 33, 34, 38 

(Graphical methods) 98 

when path of crosshead does not pass 

through center of crank-pin circle 95 

Air resistance brake 18 

Alden brake 17 

Astaticity, pendulum governors 168 

Axes, moving 200 

Axes of inertia, principal 22 

Balanced compound locomotive 121 

Balancing locomotives 116 

revolving masses Ill 

Case 2 9 

Case 3 5 

Batchelder dynamometer 115 

Belt dynamometers 113 

Brake, Alden 17 

air resistance 18 

Brakes 13 

Prony 13 

Cars, dynamometer 1 

Centrifugal tension, in rim of pulley 73 

Cone, right circular, moments of inertia 27 

solid, moments of inertia 28 

thin, hollow, moments of inertia 28 

Connecting rods of locomotives, stresses in 123 

rod, point of application of throw in line of dead points. ... 47 
point of application of throw perpendicular to line of 

dead points 50 

throw in line of dead points 47 

throw perpendicular to line of dead points 49 

Cradle, dynamometer 10 

Crank shafts, and other moving parts 132 

Critical speed 219 

deduction of f ormulae (Appendix C) 238 

Cross-armed governors 164 

Cylinder, circular, moments of inertia 25 

hollow, moments of inertia 26 

thin, hollow, moments of inertia 26, 32 

AE — i.e., fluctuation of energy 68 

Differential dynamometers 5 

Dynamometer 1 

cars 1 

243 



244 INDEX 

PAGE 

Electric dynamometers 19 

Ellipsoid of gyration (Appendix A) 231 

Emerson power scale 7 

Equivalent system of concentrated weights 105 

Euler's dynamical equations 203, 204 

kinematical equations 200 

Fan, dynamometric, Renard 18 

Flat, circular plate, moments of inertia 27 

Fluctuation of energy, AE 68 

Flywheel, acceleration when load is suddenly changed 70 

Flywheel governors 168 

approximate limits of variation of speed 178 

deduction of moment equation, Case 1 171 

General Case 172 

with one swinging weight 174 

Flywheel, moment of inertia 29 

Flywheels '. 68 

stresses in rim and rim joints due to centrifugal force 75 

velocity and displacement curves illustrated 72 

Fottinger torsion meter 12 

Gas engines, rotative effect. 86 

Governors 135 

flywheel 168 

pendulum 137 

Graduation of differential dynamometers 6 

Graphical constructions for acceleration 100 

point of application of resultant throw 106 

point of application of throw parallel to line 

of dead points 105 

velocity 98 

methods for action of reciprocating parts 98 

Harmonic motion 36 

Hartnell governor , 166 

Inertia governors 180 

dynamical treatment 196 

formulae for M , 0, and rj, etc 191 

graphical representation of moment equation 181 

oscillations under constant load 185 

moments and products of 20 

Isochronism 160, 178 

Limits of variation of speed in flywheels 69 

flywheel governors. . 178 

pendulum governors 159 

Loaded and unloaded pendulum governors 155 

Locomotive, balanced compound 121 

connecting rod, moment of inertia 30 

stresses in 123 

balancing 116 

side rods, stresses in 131 

Lombard governor 166 

Mcintosh and Seymour engine, details 58 

Meters, torsion 11 

Momental ellipsoid (Appendix A) 230 



INDEX 245 

PAGE 

Moments of inertia 20 

about any axes 20 

parallel axes 21 

various axes 25 

of circular cylinder. . 25 

of flywheel, for proper regulation 69, 70 

of locomotive connecting rod 30 

of pulley or flywheel 29 

of right circular cone 27 

of sphere 25 

of steel basket 31, 32 

of thin, hollow cone 28 

of thin, hollow cylinder 26, 32 

of two isolated weights 27 

principal 22 

Moving axes 200 

Oscillations of inertia governors, deduction of formulae (Appendix B) . . . 232 

Otto gas engine, details 57 

Parabolic pendulum governors 162 

Pendulu^m governors 137 

approximate limits of variation of speed 159 

astaticity 168 

general discussion 151 

and deduction of moment equa- 
tion 147 

Casel 141 

Case 2 143 

loaded and unloaded 155 

parabolic 162 

graphical representation of moment equation 145 

rough approximations 146 

statical discussion. 138 

simple revolving 136 

Piston and crank position when acceleration in line of dead point is zero 56 

Plate, flat, circular, moments of inertia 27 

Porter- Allen engine 57 

cards corrected for action of reciprocating parts, and 

rotative effect, diagrams 64 

Power scale, Emerson 7 

Principal axes of inertia 22 

and moments of inertia (Appendix A) 223 

moments of inertia 22 

Products of inertia 20 

about various axes 25 

Proell governor 165 

Prony brakes 13 

Pulley, centrifugal tension in rim 73 

moments of inertia 29 

stresses in arms, due to centrifugal force 85 

in rim and rim joints due to centrifugal force 75 

Reciprocating parts, action of 33, 35, 38 

Renard dynamometric fan 18 

Resistance, train 3 

Road vehicles, dynamometers 4 

Rotative effect 41 

in gas engines 86 

Rotative effect, tables 43 



OCT IS 1911 



246 INDEX 

PAGE 

Side rods of locomotives, stresses in 131 

Silver and Gay dynamometer 5 

Slotted crosshead 36 

Sphere, moment of inertia 25 

Steel basket, moments of inertia 31, 32 

Stresses in arms of pulleys 85 

locomotive connecting rods 123 

side rods of locomotives 131 

rim and rim joints of pulleys due to centrifugal force 75 

Stroke cards 33 

Tables for accelerations parallel to line of dead points 52 

for rotative effect 43 

of throw for various engines. 58, 59, 60, 61, 62, 63 

Tandem compound engine — method of proceeding illustrated 65 

Tatham dynamometer 10 

Top and gyroscope, applications in engineering 216 

— first special case 207 

second special case 211 

third special case 214 

general theory 204 

Torsion meters 11 

Total rotative effect 66 

Traction dynamometers 4 

Train resistance 3 

Transmission dynamometers 4 

Triple expansion engine, rotative effect, diagrams 66 

Van Winkle dynamometer 9 

Velocity, graphical construction 98 

Velocity and displacement curves for flywheels illustrated 72 

Webber dynamometer 5 

Weight of reciprocating parts 66 

Weights, two isolated, moments of inertia 27 



i 



